Samacheer Kalvi Class 10 Maths Solutions Chapter 8 Statistics and Probability More Ques

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Additional Questions

I. Multiple Choice Questions

 

Question 1. The range of the first 10 prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 is __________
(a) 28
(b) 29
(c) 27
Answer: (c) 27
The range is found by subtracting the smallest number from the largest number. For the given prime numbers, the largest is 29 and the smallest is 2. So, the range is \( 29 - 2 = 27 \). This quickly shows the spread of the data.
In simple words: To find the range, simply take the biggest number and subtract the smallest number from it. This shows how spread out the numbers are.

🎯 Exam Tip: Always list the numbers in order (or find the min/max) before calculating the range to avoid errors.

 

Question 2. The least value in a collection of data is 14.1. If the range of the collection is 28.4, then the greatest value of the collection is __________
(a) 42.5
(b) 43.5
(c) 42.4
(d) 42.1
Answer: (a) 42.5
We are given the smallest value (S) as 14.1 and the range (R) as 28.4. To find the largest value (L), we add the range to the smallest value. So, \( L = S + R = 14.1 + 28.4 = 42.5 \). This allows us to determine the maximum value when the minimum and spread are known.
In simple words: If you know the smallest number and how much the numbers spread out (the range), you can find the biggest number by adding them together.

🎯 Exam Tip: Remember that range is 'largest - smallest'. This means 'largest = smallest + range' and 'smallest = largest - range'.

 

Question 3. The greatest value of a collection of data is 72 and the least value is 28. Then the coefficient of range is __________
(a) 44
(b) 0.72
(c) 0.44
(d) 0.28
Answer: (c) 0.44
The coefficient of range is found by using the formula \( \frac{L-S}{L+S} \), where L is the greatest value and S is the least value. Here, \( L = 72 \) and \( S = 28 \). Plugging these into the formula, we get \( \frac{72-28}{72+28} = \frac{44}{100} = 0.44 \). This value helps compare the relative spread of different datasets.
In simple words: The coefficient of range tells us how much the data varies compared to its size. We calculate it by subtracting the smallest from the largest, and then dividing by the sum of the largest and smallest.

🎯 Exam Tip: Always be careful with the formula for coefficient of range, remembering to subtract in the numerator and add in the denominator.

 

Question 4. For a collection of 11 items, \( \Sigma x = 132 \), then the arithmetic mean is __________
(a) 11
(b) 12
(c) 14
(d) 13
Answer: (b) 12
The arithmetic mean (average) is calculated by dividing the sum of all items \( (\Sigma x) \) by the number of items \( (n) \). Here, \( \Sigma x = 132 \) and \( n = 11 \). So, the mean \( \bar{x} = \frac{132}{11} = 12 \). This is a basic measure that represents the central value of the data.
In simple words: To find the average, you add up all the numbers and then divide by how many numbers there are. This gives you a central value.

🎯 Exam Tip: Know the formula for the arithmetic mean: Sum of observations divided by the number of observations.

 

Question 5. For any collection of n items, \( \Sigma(x - \bar{x}) = \) __________
(a) \( \Sigma x \)
(b) \( \bar{x} \)
(c) (empty option)
(d) 0
Answer: (d) 0
It is a fundamental property of the arithmetic mean that the sum of the deviations of all items from their mean is always zero. This happens because the mean perfectly balances the positive and negative deviations. This property is key to understanding how the mean acts as a central balancing point.
In simple words: If you take each number and subtract the average from it, and then add all those differences together, the total will always be zero. This is a special rule about averages.

🎯 Exam Tip: Understand that the sum of deviations from the mean is always zero; this is a key concept in statistics.

 

Question 6. For any collection of n items, \( (\Sigma x) - \bar{x} = \) __________
(a) \( (n-2)\bar{x} \)
(b) \( (n - 1)\bar{x} \)
(c) 0
Answer: (b) \( (n - 1)\bar{x} \)
We know that the sum of all items \( (\Sigma x) \) is equal to \( n \) times the arithmetic mean \( (\bar{x}) \), so \( \Sigma x = n\bar{x} \). Substituting this into the expression \( \Sigma x - \bar{x} \), we get \( n\bar{x} - \bar{x} \). Factoring out \( \bar{x} \), we have \( \bar{x}(n-1) \). This relationship is useful in various statistical calculations, simplifying expressions involving sums and means.
In simple words: The total sum of numbers is the average multiplied by how many numbers there are. So, if you take the total sum and subtract the average, you get the average multiplied by one less than the number of items.

🎯 Exam Tip: Always remember the relationship \( \Sigma x = n\bar{x} \). This substitution is key in many mean-related problems.

 

Question 7. If t is the standard deviation of x, y, z, then the standard deviation of x + 5, y + 5, z + 5 is __________
(a) \( \frac{t}{3} \)
(b) \( t + 5 \)
(c) t
(d) xyz
Answer: (c) t
Adding or subtracting a constant value to every item in a dataset does not change its standard deviation. The standard deviation measures the spread of the data points around the mean, and shifting all points by the same amount doesn't change their relative spread. Therefore, if the standard deviation of x, y, z is t, then the standard deviation of x+5, y+5, z+5 remains t. This is a fundamental property of standard deviation.
In simple words: When you add or subtract the same number to all values in a set of data, the spread of the data (standard deviation) stays the same. The whole group just shifts, but it doesn't get wider or narrower.

🎯 Exam Tip: Understand that standard deviation is unaffected by changes in origin (addition/subtraction of a constant) but is affected by changes in scale (multiplication/division by a constant).

 

Question 8. If the standard deviation of a set of data is 1.6, then the variance is __________
(a) 0.4
(b) 2.56
(c) 1.96
(d) 0.04
Answer: (b) 2.56
Variance is defined as the square of the standard deviation. Given that the standard deviation (S.D.) is 1.6, the variance is \( (1.6)^2 \). Calculating this, we find \( (1.6)^2 = 2.56 \). Variance provides a measure of the spread of data points from their mean, giving squared units.
In simple words: Variance is simply the standard deviation multiplied by itself. If the standard deviation is 1.6, you multiply 1.6 by 1.6 to get the variance.

🎯 Exam Tip: Remember the direct relationship: Variance \( = (\text{Standard Deviation})^2 \) and Standard Deviation \( = \sqrt{\text{Variance}} \).

 

Question 9. If the variance of a data is 12.25, then the S.D is __________
(a) 3.5
(b) 3
(c) 2.5
(d) 3.25
Answer: (a) 3.5
The standard deviation (S.D.) is the square root of the variance. Given the variance is 12.25, we take the square root of 12.25. So, \( \text{S.D.} = \sqrt{12.25} = 3.5 \). This value gives the typical deviation of data points from the average in original units.
In simple words: To find the standard deviation, you take the square root of the variance. It's like working backward from the previous problem.

🎯 Exam Tip: Be comfortable converting between variance and standard deviation by squaring or taking the square root.

 

Question 10. Variance of the first 11 natural numbers is __________
(a) \( \sqrt{5} \)
(b) \( \sqrt{10} \)
(c) \( 5\sqrt{2} \)
(d) 10
Answer: (d) 10
The formula for the variance of the first 'n' natural numbers is \( \frac{n^{2}-1}{12} \). For the first 11 natural numbers, \( n=11 \). Substituting this into the formula gives \( \frac{11^2-1}{12} = \frac{121-1}{12} = \frac{120}{12} = 10 \). This formula is a convenient shortcut for a common statistical calculation.
In simple words: There's a quick way to find the variance of the first few counting numbers. You square the number of items, subtract 1, and then divide by 12. For 11 numbers, it comes out to 10.

🎯 Exam Tip: Memorize common formulas like the variance of the first 'n' natural numbers to save time in exams.

 

Question 11. The variance of 10, 10, 10, 10, 10 is __________
(a) 10
(b) 5
(c) 0
Answer: (c) 0
When all the values in a dataset are the same, there is no variation among them. The mean \( (\bar{x}) \) will be equal to each value. Consequently, the deviation of each value from the mean \( (d = x - \bar{x}) \) will be 0. Since there's no spread, the variance, which measures this spread, must be 0. This illustrates that variance of zero means no data variability.
In simple words: If all the numbers in a group are exactly the same, they don't spread out at all. Because there is no spread, their variance is zero.

🎯 Exam Tip: A variance of zero always indicates that all data points are identical.

 

Question 12. If the variance of 14, 18, 22, 26, 30 is 32, then the variance of 28, 36, 44, 52, 60 is __________
(a) 64
(b) 128
(c) \( 32\sqrt{2} \)
(d) 32
Answer: (b) 128
First, calculate the standard deviation (S.D.) from the given variance: \( \text{S.D.} = \sqrt{32} = 4\sqrt{2} \). Observe that the second set of numbers (28, 36, 44, 52, 60) is obtained by multiplying each number in the first set (14, 18, 22, 26, 30) by 2. When each value in a data set is multiplied by a constant (k), the new standard deviation is \( k \) times the original standard deviation. So, the new S.D. is \( 2 \times 4\sqrt{2} = 8\sqrt{2} \). Finally, the new variance is the square of the new S.D.: \( (8\sqrt{2})^2 = 64 \times 2 = 128 \). This clearly shows how scaling affects data spread.
In simple words: If you multiply every number in a data set by a certain value, the standard deviation also gets multiplied by that same value. The variance then gets multiplied by the square of that value. Here, each number is doubled, so the variance becomes four times bigger.

🎯 Exam Tip: Remember that variance is affected by changes in scale: if data is multiplied by k, variance is multiplied by \( k^2 \).

 

Question 13. The standard deviation of a collection of data is \( 2\sqrt{2} \). If each value is multiplied by 3, then the standard deviation of the new data is __________
(a) \( 4\sqrt{2} \)
(b) \( 6\sqrt{2} \)
(c) \( 9\sqrt{2} \)
Answer: (b) \( 6\sqrt{2} \)
When each value in a dataset is multiplied by a constant 'k', the new standard deviation also gets multiplied by 'k'. Here, the original standard deviation is \( 2\sqrt{2} \), and each value is multiplied by 3. Therefore, the new standard deviation is \( 3 \times 2\sqrt{2} = 6\sqrt{2} \). This property is crucial for understanding how data spread changes with scaling operations.
In simple words: If you multiply every number in a group by the same amount, the standard deviation of the new group will also be multiplied by that same amount. It means the spread grows bigger in the same way the numbers do.

🎯 Exam Tip: Know that multiplying data by a constant 'k' multiplies the standard deviation by |k|.

 

Question 14. Given \( \Sigma(x - \bar{x})^2 = 48 \), \( \bar{x} = 20 \) and \( n = 12 \). The coefficient of variation is __________
(a) 25
(b) 20
(c) 30
(d) 10
Answer: (d) 10
First, we need to find the standard deviation \( (\sigma) \). The formula for standard deviation when given the sum of squared deviations from the mean is \( \sigma = \sqrt{\frac{\Sigma(x-\bar{x})^2}{n}} \). Substituting the given values, \( \sigma = \sqrt{\frac{48}{12}} = \sqrt{4} = 2 \). Next, the coefficient of variation (C.V.) is calculated as \( \frac{\sigma}{\bar{x}} \times 100 \). With \( \sigma = 2 \) and \( \bar{x} = 20 \), we get \( \text{C.V.} = \frac{2}{20} \times 100 = \frac{1}{10} \times 100 = 10 \). This coefficient helps compare variability between datasets with different means.
In simple words: To find the coefficient of variation, first find the standard deviation by taking the square root of the sum of squared differences from the mean, divided by the number of items. Then, divide this standard deviation by the mean and multiply by 100.

🎯 Exam Tip: Remember the two-step process for coefficient of variation: first calculate standard deviation, then apply the C.V. formula.

 

Question 15. Mean and standard deviation of a data are 48 and 12 respectively. The coefficient of variation is __________
(a) 42
(b) 25
(c) 28
(d) 48
Answer: (b) 25
The coefficient of variation (C.V.) is a measure of relative variability, expressed as a percentage. It is calculated by dividing the standard deviation \( (\sigma) \) by the mean \( (\bar{x}) \) and then multiplying by 100. Given \( \sigma = 12 \) and \( \bar{x} = 48 \), we have \( \text{C.V.} = \frac{12}{48} \times 100 = \frac{1}{4} \times 100 = 25 \). This value helps compare the consistency of different datasets, regardless of their original scale.
In simple words: To find the coefficient of variation, you divide the standard deviation by the average, and then multiply by 100 to get a percentage. This helps to see how much the numbers vary compared to their average size.

🎯 Exam Tip: Coefficient of variation is particularly useful when comparing the variability of two different datasets that have different means.

 

Question 16. If \( \Phi \) is an impossible event, then P(\( \Phi \)) = __________
(a) 1
(b) \( \frac{1}{4} \)
(c) 0
(d) \( \frac{1}{2} \)
Answer: (c) 0
An impossible event is an event that cannot happen under any circumstances. By definition, the probability of such an event is always 0. For example, rolling a 7 on a standard six-sided die is an impossible event. This is a basic rule in probability theory.
In simple words: An impossible event is something that can never happen. The chance of it happening is zero.

🎯 Exam Tip: Remember the basic range of probabilities: 0 for an impossible event, and 1 for a sure event.

 

Question 17. If S is the sample space of a random experiment, then P(S) = __________
(a) 0
(b) \( \frac{1}{8} \)
(c) \( \frac{1}{2} \)
(d) 1
Answer: (d) 1
The sample space (S) represents all possible outcomes of a random experiment. Since it includes every possible outcome, the event of 'something from the sample space happening' is a certain event. The probability of a certain event is always 1. For example, when flipping a coin, getting either heads or tails is a sure event. This is another fundamental rule of probability.
In simple words: The sample space includes all possible results of an experiment. So, the chance that one of these results will happen is 1, meaning it's certain.

🎯 Exam Tip: Distinguish between impossible events (P=0) and sure events (P=1) in probability.

 

Question 18. If p is the probability of an event A, then p satisfies __________
(a) \( 0 < p < 1 \)
(b) \( 0 \le p \le 1 \)
(c) \( 0 < p < 1 \)
(d) \( 0 < p < 1 \)
Answer: (b) \( 0 \le p \le 1 \)
The probability of any event must lie between 0 and 1, inclusive. This means the probability can be 0 (for an impossible event), 1 (for a certain event), or any value in between. Probabilities are never negative or greater than 1, as they represent a proportion of outcomes. This range ensures probabilities are always physically meaningful.
In simple words: A probability number must always be between 0 and 1. It can be 0 (if something cannot happen), 1 (if it must happen), or a fraction in between.

🎯 Exam Tip: Always check that your calculated probabilities fall within the range [0, 1]. If not, there's a calculation error.

 

Question 19. Let A and B be any two events and S be the corresponding sample space. Then P(\( \bar{A} \cap B \)) = __________
(a) P(B) – P(A \( \cap \) B)
(b) P(A \( \cap \) B) – P(B)
(c) P(S)
(d) P[(A \( \cup \) B)']
Answer: (a) P(B) – P(A \( \cap \) B)
The expression \( \bar{A} \cap B \) represents the event where B occurs but A does not. This region in a Venn diagram is the part of circle B that does not overlap with circle A. Mathematically, this is equivalent to the probability of B minus the probability of the intersection of A and B, i.e., \( P(B) - P(A \cap B) \). This is a key concept in set theory and probability to identify specific non-overlapping regions.
In simple words: The chance of B happening but A not happening is the same as the chance of B happening, minus the chance of both A and B happening together. Imagine only the part of circle B that is outside circle A.

🎯 Exam Tip: Visualize the Venn diagram to understand set operations; \( \bar{A} \cap B \) represents elements that are in B but not in A.

 

Question 20. The probability that a student will score centum in mathematics is \( \frac{4}{5} \). The probability that he will not score centum is __________
(a) \( \frac{1}{5} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{3}{5} \)
(d) \( \frac{4}{5} \)
Answer: (a) \( \frac{1}{5} \)
If P(A) is the probability of an event A happening, then the probability of event A not happening (its complement, denoted as \( \bar{A} \)) is \( 1 - P(A) \). Given P(scoring centum) is \( \frac{4}{5} \), the probability of not scoring centum is \( 1 - \frac{4}{5} = \frac{1}{5} \). The sum of probabilities of an event and its complement is always 1, ensuring all outcomes are covered.
In simple words: If you know the chance of something happening, the chance of it not happening is simply 1 minus that chance. It's like saying if there's a 4/5 chance of passing, there's a 1/5 chance of not passing.

🎯 Exam Tip: Always use the complement rule \( P(\bar{A}) = 1 - P(A) \) for 'not' probabilities; it's quicker and reduces errors.

 

Question 21. If A and B are two events such that P(A) = 0.25, P(B) = 0.05 and P(A \( \cap \) B) = 0.14, then P(A \( \cup \) B) = __________
(a) 0.61
(b) 0.16
(c) 0.14
(d) 0.6
Answer: (b) 0.16
For any two events A and B, the probability of their union (A or B happening) is given by the addition rule: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Substituting the given values: \( P(A \cup B) = 0.25 + 0.05 - 0.14 = 0.30 - 0.14 = 0.16 \). This formula accounts for the overlap between events A and B, ensuring no outcome is counted twice.
In simple words: To find the chance of A or B happening, add their individual chances, then subtract the chance of both happening at the same time. This avoids counting the overlap twice.

🎯 Exam Tip: Master the addition rule for probabilities: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Pay attention to the 'minus intersection' part.

 

Question 22. There are 6 defective items in a sample of 20 items. One item is drawn at random. The probability that it is a non-defective item is __________
(a) \( \frac{1}{10} \)
(b) 0
(c) \( \frac{2}{3} \)
Answer: (a) \( \frac{7}{10} \)
The total number of items in the sample is 20. There are 6 defective items. So, the number of non-defective items is \( 20 - 6 = 14 \). The probability of drawing a non-defective item is the number of non-defective items divided by the total number of items, which is \( \frac{14}{20} \). This fraction simplifies to \( \frac{7}{10} \). This is a basic calculation of favorable outcomes over total outcomes.
In simple words: To find the chance of picking a good item, first count how many good items there are (total minus defective). Then divide the number of good items by the total number of items.

🎯 Exam Tip: Always clearly identify the number of favorable outcomes and the total number of possible outcomes before calculating probability.

 

Question 23. If A and B are mutually exclusive events and S is the sample space such that P(A) = \( \frac{1}{3} \) P(B) and S = A \( \cup \) B, then P(A) = __________
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{3}{4} \)
(d) \( \frac{3}{8} \)
Answer: (a) \( \frac{1}{4} \)
We are given that A and B are mutually exclusive events, meaning they cannot happen at the same time, so \( P(A \cap B) = 0 \). Also, the sample space S is the union of A and B, so \( P(S) = P(A \cup B) = 1 \). For mutually exclusive events, \( P(A \cup B) = P(A) + P(B) \). We are given \( P(A) = \frac{1}{3} P(B) \), which means \( P(B) = 3P(A) \). Substituting this into the union formula: \( 1 = P(A) + 3P(A) \implies 1 = 4P(A) \). Therefore, \( P(A) = \frac{1}{4} \). This calculation demonstrates combining properties of probability.
In simple words: If two events cannot happen together and cover all possibilities, their chances add up to 1. If one event's chance is a third of the other, we can use these facts to find each individual chance.

🎯 Exam Tip: Mutually exclusive events simplify the addition rule \( P(A \cup B) = P(A) + P(B) \). If they are also exhaustive (cover the entire sample space), their sum is 1.

 

Question 24. The probabilities of three mutually exclusive events A, B and C are given by \( \frac{1}{3}, \frac{1}{4} \) and \( \frac{5}{12} \). Then P(A \( \cup \) B \( \cup \) C) is __________
(a) \( \frac{19}{12} \)
(b) \( \frac{7}{12} \)
(c) 1
Answer: (c) 1
For three mutually exclusive events A, B, and C, the probability of their union (A or B or C happening) is simply the sum of their individual probabilities. This is because there is no overlap to subtract. So, \( P(A \cup B \cup C) = P(A) + P(B) + P(C) \). Substituting the given probabilities: \( \frac{1}{3} + \frac{1}{4} + \frac{5}{12} \). To add these, we find a common denominator, which is 12. So, \( \frac{4}{12} + \frac{3}{12} + \frac{5}{12} = \frac{4+3+5}{12} = \frac{12}{12} = 1 \). This indicates that these three events together cover all possible outcomes, making their union a certain event.
In simple words: If three events cannot happen at the same time, the chance of any one of them happening is found by adding up their individual chances. If this sum is 1, it means one of them is certain to happen.

🎯 Exam Tip: For mutually exclusive events, the probability of their union is the sum of their individual probabilities. Always ensure denominators are common before adding fractions.

 

Question 25. If P(A) = 0.25, P(B) = 0.50, P(A \( \cap \) B) = 0.14 then P(neither A nor B) = __________
(a) 0.39
(b) 0.25
(c) 0.11
(d) 0.24
Answer: (a) 0.39
To find the probability of neither A nor B occurring, we first find the probability of A or B occurring, \( P(A \cup B) \), using the addition rule: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.25 + 0.50 - 0.14 = 0.61 \). The event 'neither A nor B' is the complement of 'A or B', which is \( \bar{A} \cap \bar{B} \). By De Morgan's Law, this is equivalent to \( (A \cup B)' \). So, \( P(\text{neither A nor B}) = 1 - P(A \cup B) = 1 - 0.61 = 0.39 \). This is a common application of probability rules, showing how to find the probability of a compound event not happening.
In simple words: First, find the chance of A or B happening. Then, the chance of neither A nor B happening is simply 1 minus that 'A or B' chance. Think of it as 'not (A or B)'.

🎯 Exam Tip: Remember De Morgan's laws in probability: \( P(\text{neither A nor B}) = P(\bar{A} \cap \bar{B}) = P((A \cup B)') = 1 - P(A \cup B) \).

 

Question 26. A bag contains 5 black balls, 4 white balls and 3 red balls. If a ball is selected at random. The probability that it is not red is __________
(a) \( \frac{5}{12} \)
(b) \( \frac{4}{12} \)
(c) \( \frac{3}{12} \)
(d) \( \frac{3}{4} \)
Answer: (d) \( \frac{3}{4} \)
First, find the total number of balls: \( 5 + 4 + 3 = 12 \) balls. The number of red balls is 3. The probability of selecting a red ball, \( P(R) \), is \( \frac{3}{12} \). The probability that the selected ball is not red, \( P(\bar{R}) \), is the complement of \( P(R) \). So, \( P(\bar{R}) = 1 - P(R) = 1 - \frac{3}{12} = \frac{12-3}{12} = \frac{9}{12} \). This fraction simplifies to \( \frac{3}{4} \). Alternatively, you can count the non-red balls (5 black + 4 white = 9) and divide by the total (12), which gives \( \frac{9}{12} = \frac{3}{4} \).
In simple words: First, count all the balls. Then, count the balls that are *not* red. The chance of picking a non-red ball is the count of non-red balls divided by the total count.

🎯 Exam Tip: For 'not' probabilities, you can either calculate \( 1 - P(\text{event}) \) or directly count the number of outcomes that are 'not' the event.

 

Question 27. Two dice are thrown simultaneously. The probability of getting a doublet is __________
(a) \( \frac{1}{36} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{2}{3} \)
Answer: (c) \( \frac{1}{6} \)
When two dice are thrown, the total number of possible outcomes is \( 6 \times 6 = 36 \). A doublet means both dice show the same number. The possible doublets are {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}. There are 6 such outcomes. The probability of getting a doublet is the number of favorable outcomes divided by the total outcomes: \( \frac{6}{36} \), which simplifies to \( \frac{1}{6} \). This calculation is a fundamental concept in basic probability for dice rolls.
In simple words: When rolling two dice, there are 36 possible results. A 'doublet' means both dice show the same number (like two 1s or two 2s). There are 6 such doublets. So the chance is 6 out of 36, which simplifies to 1 out of 6.

🎯 Exam Tip: For dice problems, always remember the sample space size: 6 for one die, 36 for two dice, 216 for three dice.

 

Question 28. A fair die is thrown once. The probability of getting a prime or composite number is __________
(a) 1
(b) \( \frac{5}{6} \)
(c) \( \frac{1}{6} \)
Answer: (b) \( \frac{5}{6} \)
When a fair die is thrown once, the possible outcomes (sample space S) are {1, 2, 3, 4, 5, 6}. So, \( n(S) = 6 \). We need to find the probability of getting a prime or composite number. The prime numbers on a die are {2, 3, 5}. The composite numbers on a die are {4, 6}. The number 1 is neither prime nor composite. Therefore, the favorable outcomes (prime or composite) are {2, 3, 4, 5, 6}, which is 5 outcomes. The required probability is \( \frac{5}{6} \). This problem tests both probability concepts and number theory definitions.
In simple words: On a standard die, the numbers are 1, 2, 3, 4, 5, 6. Prime numbers are 2, 3, 5. Composite numbers are 4, 6. The number 1 is special and is neither. So, 5 numbers are either prime or composite. The chance is 5 out of 6.

🎯 Exam Tip: Clearly distinguish between prime, composite, and the unique properties of the number 1 (neither prime nor composite) in number-based probability questions.

 

Question 29. Probability of getting 3 heads or 3 tails in tossing a coin 3 times is __________
(a) \( \frac{1}{8} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{3}{8} \)
(d) \( \frac{1}{2} \)
Answer: (b) \( \frac{1}{4} \)
When a coin is tossed 3 times, the total number of possible outcomes (sample space S) is \( 2^3 = 8 \). These outcomes are {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. We are looking for the event of getting either 3 heads (HHH) or 3 tails (TTT). There are 2 such favorable outcomes. The probability is the number of favorable outcomes divided by the total outcomes: \( \frac{2}{8} \), which simplifies to \( \frac{1}{4} \). This is a simple application of enumerating outcomes in a sequence of independent events.
In simple words: If you flip a coin three times, there are 8 possible results. Only two of these results are 'all heads' (HHH) or 'all tails' (TTT). So the chance of either of these happening is 2 out of 8, which is 1 out of 4.

🎯 Exam Tip: For multiple coin tosses, the total number of outcomes is \( 2^n \), where 'n' is the number of tosses. List the sample space carefully to identify favorable outcomes.

 

Question 30. A card is drawn from a pack of 52 cards at random. The probability of getting neither an ace nor a king card is __________
(a) \( \frac{2}{13} \)
(b) \( \frac{11}{13} \)
(c) \( \frac{8}{13} \)
Answer: (b) \( \frac{11}{13} \)
A standard deck has 52 cards. There are 4 ace cards and 4 king cards. So, the total number of ace or king cards is \( 4 + 4 = 8 \). We want to find the probability of drawing a card that is neither an ace nor a king. This means we are interested in the remaining cards: \( 52 - 8 = 44 \) cards. The probability is the number of favorable outcomes (non-ace, non-king cards) divided by the total number of cards: \( \frac{44}{52} \). This fraction simplifies by dividing both numerator and denominator by 4, giving \( \frac{11}{13} \). This calculation involves counting specific types of cards in a deck.
In simple words: In a deck of 52 cards, there are 4 aces and 4 kings. That's 8 cards you don't want. So, there are \( 52 - 8 = 44 \) cards that are neither aces nor kings. The chance of picking one of these is 44 out of 52, which simplifies to 11 out of 13.

🎯 Exam Tip: When calculating 'neither X nor Y' probabilities, it's often easiest to find the total number of cards that are X or Y, subtract that from the total, and then find the probability of the remaining cards.

 

Question 31. The probability that a leap year will have 53 Fridays or 53 Saturdays is
(1) \( \frac{2}{7} \)
(2) \( \frac{1}{7} \)
(3) \( \frac{4}{7} \)
(4) \( \frac{3}{7} \)
Answer: (4) \( \frac{3}{7} \)
In simple words: A leap year has 366 days, which means 52 full weeks and 2 extra days. These two extra days can be any pair of consecutive days. There are 7 possible pairs for these two days. Out of these, 3 pairs include either a Friday or a Saturday. So, the probability is 3 out of 7.

🎯 Exam Tip: Remember that a leap year has 366 days (52 weeks and 2 odd days), while a non-leap year has 365 days (52 weeks and 1 odd day).

 

Question 32. The probability that a non-leap year will have 53 Sundays and 53 Mondays is
(1) \( \frac{1}{7} \)
(2) \( \frac{2}{7} \)
(3) \( \frac{3}{7} \)
(4) 0
Answer: (4) 0
In simple words: A non-leap year has 365 days, meaning 52 full weeks and 1 extra day. This single extra day can only be one day of the week, not two. For a non-leap year to have 53 Sundays and 53 Mondays, that one extra day would need to be both Sunday and Monday at the same time, which is not possible. Therefore, the probability is 0.

🎯 Exam Tip: An event that cannot possibly happen is called an impossible event, and its probability is always 0.

 

Question 33. The probability of selecting a queen of hearts when a card is drawn from a pack of 52 playing cards is
(1) \( \frac{1}{52} \)
(2) \( \frac{16}{52} \)
(3) \( \frac{1}{13} \)
(4) \( \frac{1}{26} \)
Answer: (1) \( \frac{1}{52} \)
In simple words: In a standard deck of 52 playing cards, there is only one card that is the Queen of Hearts. Since we are drawing one card from the total 52 cards, and each card has an equal chance, the probability of picking that specific card is 1 divided by 52.

🎯 Exam Tip: Always identify the total number of possible outcomes (sample space) and the number of favorable outcomes for the specific event.

 

Question 34. Probability of sure event is
(1) 1
(2) 0
(3) 100
(4) 0.1
Answer: (1) 1
In simple words: A sure event is something that is absolutely certain to happen. For example, if you say the sun will rise tomorrow, that's a sure event. When an event is guaranteed to occur, its probability is 1. This means it has a 100% chance of happening.

🎯 Exam Tip: The probability of any event always lies between 0 (impossible event) and 1 (sure event), inclusive.

 

Question 35. The outcome of a random experiment result in either success or failure. If the probability of success is twice the probability of failure, then the probability of success is
(1) \( \frac{1}{3} \)
(2) \( \frac{2}{3} \)
(3) 1
(4) 0
Answer: (2) \( \frac{2}{3} \)
In simple words: If an experiment can only result in success or failure, then the chance of success plus the chance of failure must add up to 1. The problem says success is twice as likely as failure. If you divide the total probability (1) into three parts, success takes two parts and failure takes one part. So, the probability of success is \( \frac{2}{3} \).

🎯 Exam Tip: For events that are mutually exclusive and exhaustive (meaning only one can happen and one must happen), the sum of their probabilities is always 1.

 

II. Answer the following questions.

 

Question 1. Find the range and the coefficient of range of the following data.

Answer:
Largest value (L) = 690
Smallest value (S) = 610
Range R = L - S = \( 690 - 610 = 80 \)
Coefficient of range = \( \frac{L-S}{L+S}=\frac{690-610}{690+610}=\frac{80}{1300} \)
\( = 0.06 \)
In simple words: To find the range, we simply take the highest income value and subtract the lowest income value, which gives us 80. To find the coefficient of range, we use a formula: (Largest value - Smallest value) divided by (Largest value + Smallest value). This calculation shows how spread out the data is relative to its average size.

🎯 Exam Tip: The range gives the absolute difference, while the coefficient of range gives a relative measure, useful for comparing variability across datasets with different scales.

 

Question 2. Two dice are thrown simultaneously. What is the probability that
(i) 5 will not come up on either of them
(ii) 5 will come up at both dice
Answer:
The total sample space (S) for two dice thrown simultaneously is:
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
n(S) = 36
(i) Let A be the event of getting 5 on either of them.
A = {(1,5), (2,5), (3,5), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,5)}
n(A) = 11
P(A) = \( \frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{11}{36} \)
The probability that 5 will not come up on either of them = \( 1 - P(A) \)
\( = 1-\frac{11}{36}=\frac{36-11}{36}=\frac{25}{36} \)
(ii) Let B be the event of getting 5 will come up at both dice.
B = {(5,5)}
n(B) = 1
P(B) = \( \frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{1}{36} \)
In simple words: When two dice are rolled, there are 36 possible outcomes. (i) To find the probability that 5 does not appear on either die, we first count all outcomes where 5 *does* appear. There are 11 such outcomes. So, the probability of 5 appearing is \( \frac{11}{36} \). The probability of it *not* appearing is \( 1 - \frac{11}{36} = \frac{25}{36} \). (ii) To find the probability that 5 appears on *both* dice, we look for the specific outcome (5,5). There is only 1 such outcome, so the probability is \( \frac{1}{36} \).

🎯 Exam Tip: Clearly listing the sample space or using the multiplication rule for independent events can help avoid errors in probability calculations for multiple dice.

 

Question 3. The king, Queen and Jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards,
(i) a card of clubs
(ii) a queen of diamond
Answer:
Total cards in a deck = 52
Number of cards removed (King, Queen, Jack of clubs) = 3
Remaining cards (Sample space S) = \( 52 - 3 = 49 \)
n(S) = 49
(i) Let A be the event of getting a card of clubs.
Original number of clubs = 13
Number of clubs removed = 3 (K, Q, J of clubs)
Number of clubs remaining n(A) = \( 13 - 3 = 10 \)
P(A) = \( \frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{10}{49} \)
(ii) Let B be the event of getting a queen of diamond.
Number of queen of diamonds n(B) = 1 (Queen of Diamonds was not removed)
P(B) = \( \frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{1}{49} \)
In simple words: When three specific club cards (King, Queen, Jack) are taken out of a 52-card deck, the total number of cards left is 49. This new total is our sample space. (i) There were 13 club cards, but 3 were removed, so 10 club cards are left. The chance of drawing a club is 10 out of 49. (ii) There is only one Queen of Diamonds in a deck, and it was not among the cards removed. So, the chance of drawing a Queen of Diamonds from the 49 cards left is 1 out of 49.

🎯 Exam Tip: Always adjust both the total number of outcomes (sample space) and the number of favorable outcomes if cards are removed or added to the deck.

 

Question 4. The standard deviation of 20 observations is \( \sqrt{5} \). If each observation is multiplied by 2, find the standard deviation and variance of the resulting observations.
Answer:
Given standard deviation of 20 observations \( = \sqrt{5} \)
If each observation is multiplied by 2:
New standard deviation = \( 2 \times \sqrt{5} = 2\sqrt{5} \)
New variance = \( (2\sqrt{5})^2 \)
\( = 4 \times 5 = 20 \)
In simple words: When every single data point in a set is multiplied by a certain number, the standard deviation also gets multiplied by that exact same number. So, if the original standard deviation was \( \sqrt{5} \) and each observation is multiplied by 2, the new standard deviation becomes \( 2\sqrt{5} \). The variance is found by squaring the standard deviation. So, the new variance is \( (2\sqrt{5})^2 \), which equals 20.

🎯 Exam Tip: Remember that adding or subtracting a constant to all observations does not change the standard deviation or variance, but multiplying or dividing does.

 

Question 5. Calculate the variance standard deviation of the following data 38, 40, 34, 31, 28, 26, 34.
Answer:
Arrange the given data in ascending order: 26, 28, 31, 34, 34, 38, 40
Assumed mean (A) = 34

The number of observations (n) = 7
Variance \( = \frac{\Sigma d_i^2}{n} - (\frac{\Sigma d_i}{n})^2 \)
\( = \frac{161}{7} - (\frac{-7}{7})^2 \)
\( = 23 - (-1)^2 \)
\( = 23 - 1 = 22 \)
Standard deviation \( (\sigma) = \sqrt{\text{Variance}} \)
\( = \sqrt{22} \)
\( \approx 4.69 \)
In simple words: First, we sort the data and pick an assumed mean, which is 34 here. Then, for each number, we find its difference from the assumed mean and square that difference. We add up all these differences and squared differences. Using the formula for variance, which involves the sum of squared differences and the sum of differences, we find the variance to be 22. The standard deviation is simply the square root of this variance, which is about 4.69.

🎯 Exam Tip: Arranging data in ascending order and choosing a central value as the assumed mean can simplify calculations for grouped or ungrouped data.

 

Question 6. Mean of 100 items is 48 and their standard deviation is 10. Find the sum of all the items and the sum of the squares of all items.
Answer:
Given:
Number of items (n) = 100
Mean \( (\bar{x}) = 48 \)
Standard deviation \( (\sigma) = 10 \)
Sum of all items \( (\Sigma x) \):
Mean \( \bar{x} = \frac{\Sigma x}{n} \)
\( 48 = \frac{\Sigma x}{100} \)
\( \Sigma x = 48 \times 100 = 4800 \)
Sum of the squares of all items \( (\Sigma x^2) \):
Variance \( (\sigma^2) = (\text{Standard deviation})^2 = 10^2 = 100 \)
Variance \( (\sigma^2) = \frac{\Sigma x^2}{n} - (\frac{\Sigma x}{n})^2 \)
\( \implies 100 = \frac{\Sigma x^2}{100} - (48)^2 \)
\( 100 = \frac{\Sigma x^2}{100} - 2304 \)
\( \implies \frac{\Sigma x^2}{100} = 100 + 2304 \)
\( \frac{\Sigma x^2}{100} = 2404 \)
\( \Sigma x^2 = 2404 \times 100 = 240400 \)
In simple words: We are given the average and how spread out the data is for 100 items. To find the sum of all items, we multiply the mean by the number of items: \( 48 \times 100 = 4800 \). To find the sum of the squares of all items, we use the variance formula. Since standard deviation is 10, variance is \( 10^2 = 100 \). We put the numbers into the variance formula and solve for the sum of squares, which turns out to be 240400.

🎯 Exam Tip: Understand the relationships between mean, standard deviation, variance, sum of items, and sum of squares; they are all connected through fundamental statistical formulas.

 

Question 7. If n = 10, \( \bar{x} \) = 12 and \( \Sigma x^2 \) = 1530, then calculate the coefficient of variation.
Answer:
Given:
Number of items (n) = 10
Mean \( (\bar{x}) = 12 \)
Sum of squares \( (\Sigma x^2) = 1530 \)
First, calculate the standard deviation \( (\sigma) \):
Standard deviation \( (\sigma) = \sqrt{\frac{\Sigma x^2}{n} - (\bar{x})^2} \)
\( = \sqrt{\frac{1530}{10} - (12)^2} \)
\( = \sqrt{153 - 144} \)
\( = \sqrt{9} \)
\( \sigma = 3 \)
Now, calculate the coefficient of variation (C.V.):
Coefficient of variation = \( \frac{\sigma}{\bar{x}} \times 100 \)
\( = \frac{3}{12} \times 100 \)
\( = \frac{1}{4} \times 100 \)
\( = 25 \)
In simple words: To find the coefficient of variation, we first need to calculate the standard deviation. We use the formula \( \sigma = \sqrt{\frac{\Sigma x^2}{n} - (\bar{x})^2} \) with the given values for n, mean, and sum of squares. This gives us a standard deviation of 3. Then, we apply the formula for coefficient of variation, which is \( \frac{\sigma}{\bar{x}} \times 100 \), resulting in 25.

🎯 Exam Tip: The coefficient of variation is a dimensionless quantity, making it useful for comparing the variability of different datasets without being affected by the units of measurement.

 

Question 8. If the coefficient of variation of a collection of data is 57 and its standard deviation is 6.84, then find the mean.
Answer:
Given:
Coefficient of variation (C.V.) = 57
Standard deviation \( (\sigma) = 6.84 \)
Formula for Coefficient of variation:
C.V. \( = \frac{\sigma}{\bar{x}} \times 100 \)
Substitute the given values:
\( 57 = \frac{6.84}{\bar{x}} \times 100 \)
Now, solve for the mean \( (\bar{x}) \):
\( \bar{x} = \frac{6.84 \times 100}{57} \)
\( \bar{x} = \frac{684}{57} \)
\( \bar{x} = 12 \)
The arithmetic mean is 12.
In simple words: We are given the coefficient of variation and the standard deviation, and we need to find the mean. We use the formula for coefficient of variation, which is \( \text{C.V.} = \frac{\sigma}{\bar{x}} \times 100 \). We rearrange this formula to solve for the mean: \( \bar{x} = \frac{\sigma \times 100}{\text{C.V.}} \). Plugging in the given numbers, we calculate the mean to be 12.

🎯 Exam Tip: Always remember to correctly rearrange the formula to solve for the unknown variable, ensuring all units are consistent before calculation.

 

Question 9. Find the standard deviation and the variance of first 23 natural numbers?
Answer:
To find the standard deviation of the first 'n' natural numbers, we use the formula:
Standard deviation \( (\sigma) = \sqrt{\frac{n^2-1}{12}} \)
Here, n = 23 (first 23 natural numbers).
Substitute n = 23 into the formula:
\( \sigma = \sqrt{\frac{23^2-1}{12}} \)
\( = \sqrt{\frac{529-1}{12}} \)
\( = \sqrt{\frac{528}{12}} \)
\( = \sqrt{44} \)
\( \approx 6.63 \)
Now, calculate the variance:
Variance \( = (\text{Standard deviation})^2 \)
\( = (\sqrt{44})^2 \)
\( = 44 \)
In simple words: There's a special shortcut formula to find the standard deviation of a series of natural numbers. For the first 'n' natural numbers, the standard deviation is \( \sqrt{\frac{n^2-1}{12}} \). For the first 23 numbers, this calculates to \( \sqrt{44} \), which is about 6.63. The variance is simply the square of the standard deviation, so it is 44.

🎯 Exam Tip: Memorizing specific formulas for common series like natural numbers can save significant time on exams and prevent lengthy individual calculations.

 

Question 10. Find the coefficient of variation of the following data: 18, 20, 15, 12, 25.
Answer:
Let us calculate the arithmetic mean (A.M.) of the given data.
\( \bar{x} = \frac{12+15+18+20+25}{5} = \frac{90}{5} = 18 \)

Number of items (n) = 5
Standard deviation \( (\sigma) = \sqrt{\frac{\Sigma d^2}{n}} \)
\( = \sqrt{\frac{98}{5}} \)
\( = \sqrt{19.6} \)
\( \approx 4.428 \)
Now, calculate the coefficient of variation (C.V.):
Coefficient of variation \( = \frac{\sigma}{\bar{x}} \times 100 \)
\( = \frac{4.428}{18} \times 100 \)
\( = \frac{442.8}{18} \)
\( \approx 24.6 \)
In simple words: First, we find the average of the numbers, which is 18. Then, for each number, we find how much it differs from the average and square that difference. The sum of these squared differences is 98. We divide this sum by the number of data points (5) and take the square root to get the standard deviation, which is about 4.428. Finally, to get the coefficient of variation, we divide the standard deviation by the average and multiply by 100, giving us roughly 24.6.

🎯 Exam Tip: The coefficient of variation is a percentage and helps in comparing the relative variability of data sets, even if they have different units or means.

 

Question 11. Three rotten eggs are mixed with 12 good ones. One egg is chosen at random. What is the probability of choosing a rotten egg?
Answer:
Number of good eggs = 12
Number of rotten eggs = 3
Total number of eggs = \( 12 + 3 = 15 \)
The total possible outcomes (sample space n(S)) = 15
Let A be the event of choosing a rotten egg.
Number of favorable outcomes n(A) = 3
Probability P(A) = \( \frac{n(\mathrm{A})}{n(\mathrm{S})} \)
\( = \frac{3}{15} \)
\( = \frac{1}{5} \)
The probability of choosing a rotten egg is \( \frac{1}{5} \).
In simple words: We have 3 rotten eggs and 12 good eggs, which means there are 15 eggs in total. If we pick one egg without looking, there are 15 possible outcomes. Since 3 of these outcomes are rotten eggs, the chance of picking a rotten egg is 3 out of 15. This simplifies to 1 out of 5.

🎯 Exam Tip: When calculating probability, always ensure you have correctly identified both the number of favorable outcomes and the total number of possible outcomes.

 

Question 12. Two coins are tossed together. What is the probability of getting at most one head?
Answer:
When two coins are tossed, the sample space (S) is:
S = {(H, H), (H, T), (T, H), (T, T)}
The total number of outcomes n(S) = 4
Let A be the event of getting at most one head. This means getting either one head or zero heads.
A = {(H, T), (T, H), (T, T)}
The number of favorable outcomes n(A) = 3
Probability P(A) = \( \frac{n(\mathrm{A})}{n(\mathrm{S})} \)
\( = \frac{3}{4} \)
The probability is \( \frac{3}{4} \).
In simple words: When two coins are flipped, there are four possible results: two heads (HH), one head and one tail (HT or TH), or two tails (TT). "At most one head" means we can have one head or no heads. The results that fit this are HT, TH, and TT. There are 3 such results. So, the chance is 3 out of the 4 total possible results.

🎯 Exam Tip: "At most" means "less than or equal to," so ensure you include outcomes that have zero occurrences of the specified item.

 

Question 13. A number is selected at random from integers 1 to 100. Find the probability that it is
(i) a perfect square
(ii) not a perfect cube.
Answer:
The sample space (S) = {1, 2, 3, ..., 100}
The total number of outcomes n(S) = 100
(i) Let A be the event of getting a perfect square.
The perfect squares between 1 and 100 are:
A = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
The number of favorable outcomes n(A) = 10
Probability P(A) = \( \frac{n(\mathrm{A})}{n(\mathrm{S})} \)
\( = \frac{10}{100} \)
\( = \frac{1}{10} \)
(ii) Let B be the event of getting a perfect cube.
The perfect cubes between 1 and 100 are:
B = {1, 8, 27, 64}
The number of favorable outcomes n(B) = 4
Probability P(B) = \( \frac{n(\mathrm{B})}{n(\mathrm{S})} \)
\( = \frac{4}{100} \)
\( = \frac{1}{25} \)
The probability that the selected number is not a perfect cube is P(\( \bar{B} \)).
P(\( \bar{B} \)) = \( 1 - P(B) \)
\( = 1 - \frac{1}{25} \)
\( = \frac{25-1}{25} \)
\( = \frac{24}{25} \)
In simple words: When a number is picked from 1 to 100, there are 100 total possibilities. (i) Perfect squares are numbers like 1, 4, 9, and so on, up to 100. There are 10 such numbers. So, the chance of picking a perfect square is 10 out of 100, which is \( \frac{1}{10} \). (ii) Perfect cubes are numbers like 1, 8, 27, and 64, up to 100. There are 4 such numbers. The chance of picking a perfect cube is \( \frac{4}{100} = \frac{1}{25} \). The chance of *not* picking a perfect cube is \( 1 - \frac{1}{25} = \frac{24}{25} \).

🎯 Exam Tip: For "not" events (complementary events), calculate the probability of the event happening and subtract it from 1.

 

Question 14. Three dice are thrown simultaneously. Find the probability of getting the same number on all the three dice.
Answer:
When three dice are thrown simultaneously, each die has 6 possible outcomes. So, the total number of possible outcomes in the sample space is:
n(S) = \( 6 \times 6 \times 6 = 6^3 = 216 \)
Let A be the event of getting the same number on all three dice.
The favorable outcomes are:
A = {(1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6)}
The number of favorable outcomes n(A) = 6
Probability P(A) = \( \frac{n(A)}{n(S)} \)
\( = \frac{6}{216} \)
\( = \frac{1}{36} \)
The probability of getting the same number on all three dice is \( \frac{1}{36} \).
In simple words: When you roll three dice, each die can land on one of six numbers. So, the total number of different results you can get is \( 6 \times 6 \times 6 = 216 \). We want to find the chance that all three dice show the exact same number. These results are (1,1,1), (2,2,2), all the way to (6,6,6). There are 6 such results. So, the probability is 6 out of 216, which simplifies to \( \frac{1}{36} \).

🎯 Exam Tip: For independent events like rolling multiple dice, the total number of outcomes is found by multiplying the number of outcomes for each individual event.

 

Question 15. If P(A) = \( \frac{1}{2} \), P(B) = \( \frac{7}{10} \), P(A∪B) = 1, Find
(i) P(A ∩ B)
(ii) P(A' ∪ B')
Answer:
Given:
P(A) = \( \frac{1}{2} \)
P(B) = \( \frac{7}{10} \)
P(A∪B) = 1
(i) To find P(A ∩ B), use the Addition Theorem for probabilities:
P(A∪B) = P(A) + P(B) - P(A ∩ B)
Rearranging the formula:
P(A ∩ B) = P(A) + P(B) - P(A∪B)
\( = \frac{1}{2} + \frac{7}{10} - 1 \)
To add and subtract these fractions, find a common denominator, which is 10:
\( = \frac{5}{10} + \frac{7}{10} - \frac{10}{10} \)
\( = \frac{5+7-10}{10} \)
\( = \frac{12-10}{10} \)
\( = \frac{2}{10} = \frac{1}{5} \)
So, P(A ∩ B) = \( \frac{1}{5} \).
(ii) To find P(A' ∪ B'), use De Morgan's Law, which states that A' ∪ B' = \( \overline{A \cap B} \).
This means P(A' ∪ B') is the probability of the complement of (A ∩ B).
P(A' ∪ B') = \( 1 - P(A \cap B) \)
Using the value from part (i):
\( = 1 - \frac{1}{5} \)
\( = \frac{5-1}{5} \)
\( = \frac{4}{5} \)
So, P(A' ∪ B') = \( \frac{4}{5} \).
In simple words: We are given the chances of two things, A and B, happening, and the chance of either A or B (or both) happening. (i) To find the chance that both A and B happen, we use a formula: add the chances of A and B, then subtract the chance of A or B (or both). This gives us \( \frac{1}{5} \). (ii) To find the chance that "not A or not B" happens, we use a special rule that says this is the same as "not (A and B)". So, we take 1 and subtract the chance that "A and B" both happen, which we found to be \( \frac{1}{5} \). This gives us \( \frac{4}{5} \).

🎯 Exam Tip: De Morgan's Laws are critical for simplifying complex probability expressions involving complements of unions or intersections.

 

III. Answer the following questions.

 

Question 1. The mean of the following frequency distribution is 53 and the sum of all frequencies is 100. compute the missing frequencies \( f_1 \) and \( f_2 \).

Answer:
To find the missing frequencies \( f_1 \) and \( f_2 \), we first set up a frequency distribution table to calculate the sum of frequencies (\( \Sigma f_i \)) and the sum of the product of frequency and mid-value (\( \Sigma f_i x_i \)).

*Correction: The OCR's table calculation had 80-100 frequency as 17 for sum in calculation, but original table shows 7. I will use 7 to calculate sum of f_i and f_i x_i. Recalculating sum of f_i: \( 15 + f_1 + 21 + f_2 + 7 = 43 + f_1 + f_2 \). Recalculating sum of f_i x_i: \( 150 + 30f_1 + 1050 + 70f_2 + (7 \times 90) = 150 + 30f_1 + 1050 + 70f_2 + 630 = 1830 + 30f_1 + 70f_2 \). The source has `Σf₁ = 53 +11 +12` and `Σfx₁ = 2730 + 30 f₁ +70 f2`. This implies different values for the last row than the input table. The input table has 7 for 80-100 freq, not 17. The given `Σfx₁` also sums `150+1050+1530 = 2730`. This means the `1530` must come from `17 * 90` if the freq was 17. Following IRON RULE 6, I must use the source's calculation values for the sum, even if they don't exactly match the input table's last row (it implicitly corrected 7 to 17 to get 1530 as the last fx value and 53 as sum of known frequencies). I will use the sums given in the solution: `Σf_i = 53 + f_1 + f_2` and `Σf_i x_i = 2730 + 30 f_1 + 70 f_2`.
Given, total sum of frequencies \( \Sigma f_i = 100 \)
From our sum: \( 53 + f_1 + f_2 = 100 \)
\( \implies f_1 + f_2 = 100 - 53 \)
\( f_1 + f_2 = 47 \) ...(1)

Given mean \( \bar{x} = 53 \)
The formula for mean is \( \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} \)
\( 53 = \frac{2730 + 30 f_1 + 70 f_2}{100} \)
Multiply both sides by 100:
\( 53 \times 100 = 2730 + 30 f_1 + 70 f_2 \)
\( 5300 = 2730 + 30 f_1 + 70 f_2 \)
Rearrange the equation:
\( 5300 - 2730 = 30 f_1 + 70 f_2 \)
\( 2570 = 30 f_1 + 70 f_2 \)
Divide the entire equation by 10 to simplify:
\( 257 = 3 f_1 + 7 f_2 \) ...(2)

Now we have a system of two linear equations:
1) \( f_1 + f_2 = 47 \)
2) \( 3 f_1 + 7 f_2 = 257 \)

From equation (1), express \( f_1 \) in terms of \( f_2 \):
\( f_1 = 47 - f_2 \)
Substitute this expression for \( f_1 \) into equation (2):
\( 3(47 - f_2) + 7 f_2 = 257 \)
\( 141 - 3 f_2 + 7 f_2 = 257 \)
\( 141 + 4 f_2 = 257 \)
\( 4 f_2 = 257 - 141 \)
\( 4 f_2 = 116 \)
\( f_2 = \frac{116}{4} \)
\( f_2 = 29 \)

Substitute the value of \( f_2 \) back into equation (1) to find \( f_1 \):
\( f_1 + 29 = 47 \)
\( f_1 = 47 - 29 \)
\( f_1 = 18 \)
Thus, the missing frequencies are \( f_1 = 18 \) and \( f_2 = 29 \).
In simple words: We need to find two missing numbers, \( f_1 \) and \( f_2 \), in a table of frequencies, given the average (mean) and the total sum of all frequencies. First, we use the total frequency to make one equation involving \( f_1 \) and \( f_2 \). Then, we calculate the mid-point for each group and use the mean formula to create a second equation. With two equations, we solve them like a puzzle to find that \( f_1 \) is 18 and \( f_2 \) is 29.

🎯 Exam Tip: Always double-check your arithmetic when solving systems of equations, especially after substitution, as a small error can lead to incorrect final frequencies.

 

Question 2. Calculate the standard deviation of the following data.

Answer:
We will use the assumed mean method to calculate the standard deviation.
Let Assumed mean (A) = 13

The formula for standard deviation \( (\sigma) \) using the assumed mean method is:
\( \sigma = \sqrt{\frac{\Sigma fd^2}{\Sigma f} - (\frac{\Sigma fd}{\Sigma f})^2} \)
Substitute the sums from the table:
\( \sigma = \sqrt{\frac{2000}{50} - (\frac{10}{50})^2} \)
\( \sigma = \sqrt{40 - (0.2)^2} \)
\( \sigma = \sqrt{40 - 0.04} \)
\( \sigma = \sqrt{39.96} \)
\( \sigma \approx 6.32 \)
The standard deviation is approximately 6.32.
In simple words: To find how spread out the numbers are, we use a method where we pick an "assumed mean" (here, 13) to make calculations simpler. We find the difference of each number from this assumed mean, multiply it by its frequency, and then also square these differences and multiply by frequency. We add up all these values. Then, we use a special formula that combines these sums to calculate the standard deviation, which comes out to be about 6.32.

🎯 Exam Tip: When using the assumed mean method, ensure the sum of deviations (`Σfd`) and sum of squared deviations (`Σfd²`) are calculated correctly for accurate results.

 

Question 3. The time (in seconds) taken by a group to walk across a pedestrian crossing is given in the table below.
Calculate the variance and standard deviation of the data.

Answer:
To calculate the variance and standard deviation for this grouped data, we use the step deviation method.
Let Assumed mean (A) = 17.5
Class width (c) = 5

The formula for Variance \( (\sigma^2) \) using the step deviation method is:
\( \sigma^2 = \left(\frac{\Sigma fd^2}{\Sigma f} - \left(\frac{\Sigma fd}{\Sigma f}\right)^2\right) \times c^2 \)
Substitute the sums from the table:
\( \sigma^2 = \left(\frac{80}{50} - \left(\frac{18}{50}\right)^2\right) \times 5^2 \)
\( = (1.6 - (0.36)^2) \times 25 \)
\( = (1.6 - 0.1296) \times 25 \)
\( = 1.4704 \times 25 \)
\( \sigma^2 = 36.76 \)
Now, calculate the Standard deviation \( (\sigma) \):
\( \sigma = \sqrt{\sigma^2} = \sqrt{36.76} \)
\( \sigma \approx 6.063 \)
The Variance is 36.76, and the Standard deviation is approximately 6.06.
In simple words: To find how spread out the data is for these time intervals, we use a method where we pick an average value (17.5 seconds) and a step size (5 seconds). We then calculate a simple deviation for each interval, multiply it by the number of people, and square it. We sum these up. Using specific formulas for variance and standard deviation that include these sums and the step size, we find the variance to be 36.76 and the standard deviation to be about 6.06.

🎯 Exam Tip: For grouped frequency data with equal class intervals, the step deviation method simplifies computations by working with smaller, scaled deviations.

 

Question 4. The mean and standard deviation of 20 items are found to be 10 and 2 respectively. At the time of checking it was found that an item 12 was wrongly entered as 8. Calculate the correct mean and standard deviation.
Answer:
Given:
Number of items (n) = 20
Incorrect mean \( (\bar{x}_{\text{incorrect}}) = 10 \)
Incorrect standard deviation \( (\sigma_{\text{incorrect}}) = 2 \)
Wrong item = 8, Correct item = 12

Step 1: Calculate the corrected mean.
From the incorrect mean, find the incorrect sum of items:
\( \bar{x}_{\text{incorrect}} = \frac{\Sigma x_{\text{incorrect}}}{n} \)
\( 10 = \frac{\Sigma x_{\text{incorrect}}}{20} \)
\( \Sigma x_{\text{incorrect}} = 10 \times 20 = 200 \)

Now, correct the sum of items:
\( \Sigma x_{\text{correct}} = \Sigma x_{\text{incorrect}} - (\text{Wrong item}) + (\text{Correct item}) \)
\( = 200 - 8 + 12 = 204 \)

Calculate the corrected mean:
\( \bar{x}_{\text{correct}} = \frac{\Sigma x_{\text{correct}}}{n} = \frac{204}{20} = 10.2 \)

Step 2: Calculate the corrected standard deviation.
From the incorrect standard deviation, find the incorrect sum of squares \( (\Sigma x^2_{\text{incorrect}}) \):
Variance \( (\sigma^2_{\text{incorrect}}) = \frac{\Sigma x^2_{\text{incorrect}}}{n} - (\bar{x}_{\text{incorrect}})^2 \)
\( 2^2 = \frac{\Sigma x^2_{\text{incorrect}}}{20} - (10)^2 \)
\( 4 = \frac{\Sigma x^2_{\text{incorrect}}}{20} - 100 \)
\( \implies \frac{\Sigma x^2_{\text{incorrect}}}{20} = 4 + 100 = 104 \)
\( \Sigma x^2_{\text{incorrect}} = 104 \times 20 = 2080 \)

Now, correct the sum of squares:
\( \Sigma x^2_{\text{correct}} = \Sigma x^2_{\text{incorrect}} - (\text{Wrong item})^2 + (\text{Correct item})^2 \)
\( = 2080 - 8^2 + 12^2 \)
\( = 2080 - 64 + 144 \)
\( = 2160 \)

Finally, calculate the corrected standard deviation:
\( \sigma_{\text{correct}} = \sqrt{\frac{\Sigma x^2_{\text{correct}}}{n} - (\bar{x}_{\text{correct}})^2} \)
\( = \sqrt{\frac{2160}{20} - (10.2)^2} \)
\( = \sqrt{108 - 104.04} \)
\( = \sqrt{3.96} \)
\( \approx 1.99 \)

(i) Corrected mean = 10.2
(ii) Corrected Standard deviation = 1.99
In simple words: We start with the given average and spread (standard deviation) for 20 numbers. We find out that one number was written down wrong. First, we correct the sum of all numbers by taking out the wrong number and putting in the right one. This gives us the new, correct average. Second, we do a similar process for the sum of the squares of the numbers. We use the variance formula to get the original sum of squares, then correct it by taking out the square of the wrong number and adding the square of the right number. Finally, using this new sum of squares and the correct average, we calculate the correct standard deviation.

🎯 Exam Tip: When correcting errors in statistical data, always remember to adjust the sum of observations for the mean and the sum of squares of observations for the standard deviation before recalculating.

 

Question 5. Calculate the coefficient of variation of the following data: 20, 18, 32, 24, 26.
Answer: First, we arrange the data in increasing order: 18, 20, 24, 26, 32. The number of items \( n = 5 \). We will use an assumed mean, \( A = 24 \), for easier calculation. This helps in simplifying calculations with larger numbers.

The arithmetic mean \( \bar{x} = \frac{\Sigma x}{n} = \frac{120}{5} = 24 \).
The standard deviation \( (\sigma) \) is calculated using the formula:
\[ \sigma = \sqrt{\frac{\Sigma d^2}{n} - \left(\frac{\Sigma d}{n}\right)^2} \] Now, substitute the values we found:
\[ \sigma = \sqrt{\frac{120}{5} - \left(\frac{0}{5}\right)^2} \]
\[ \sigma = \sqrt{24 - 0} \]
\[ \sigma = \sqrt{24} \]
\[ \sigma \approx 4.8989 \approx 4.9 \] The coefficient of variation (C.V.) is found using the formula:
\[ C.V. = \frac{\sigma}{\bar{x}} \times 100 \]
\[ C.V. = \frac{4.9}{24} \times 100 \]
\[ C.V. = 0.20416 \times 100 \]
\[ C.V. \approx 20.42 \]
In simple words: First, find the average of the numbers. Then, calculate how spread out the numbers are from the average. Finally, use these two values to find the coefficient of variation, which tells us the relative spread of the data.

🎯 Exam Tip: Remember to correctly identify the number of items (n) and use the assumed mean method for large datasets to simplify calculations, paying attention to the signs of \(d_i\).

 

Question 6. The marks scored by two students A, B in a class are given below. Who is more consistent?
Answer: To find out which student is more consistent, we need to calculate the coefficient of variation for each student. A lower coefficient of variation indicates greater consistency. **For Student A:** The marks for Student A are 58, 51, 60, 65, 66. The mean of Student A's marks \( \bar{x} = 60 \) (given). We use the mean to find the deviations \( d_i \) and squared deviations \( d_i^2 \). This helps us measure the spread of marks.

The standard deviation \( (\sigma) \) is found using the formula:
\[ \sigma = \sqrt{\frac{\Sigma d_i^2}{n} - \left(\frac{\Sigma d_i}{n}\right)^2} \] Substitute the values for Student A:
\[ \sigma = \sqrt{\frac{146}{5} - \left(\frac{0}{5}\right)^2} \]
\[ \sigma = \sqrt{29.2 - 0} \]
\[ \sigma = \sqrt{29.2} \approx 5.4037 \approx 5.4 \] The coefficient of variation (C.V.) for Student A is:
\[ C.V. = \frac{\sigma}{\bar{x}} \times 100 \]
\[ C.V. = \frac{5.4}{60} \times 100 = 0.09 \times 100 = 9 \] **For Student B:** The marks for Student B are 56, 87, 88, 46, 43. The mean of Student B's marks \( \bar{x} = 64 \) (given). We calculate the deviations and squared deviations for Student B:The standard deviation \( (\sigma) \) for Student B:
\[ \sigma = \sqrt{\frac{1934}{5} - \left(\frac{0}{5}\right)^2} \]
\[ \sigma = \sqrt{386.8 - 0} \]
\[ \sigma = \sqrt{386.8} \approx 19.667 \approx 19.67 \] The coefficient of variation (C.V.) for Student B is:
\[ C.V. = \frac{\sigma}{\bar{x}} \times 100 \]
\[ C.V. = \frac{19.67}{64} \times 100 \]
\[ C.V. \approx 0.30734 \times 100 \approx 30.73 \] Since Student A has a lower coefficient of variation (9) compared to Student B (30.73), Student A is more consistent. A smaller coefficient of variation means the data points are clustered more closely around the mean.
In simple words: We calculated a "consistency score" for both students. Student A's score was much lower than Student B's. This means Student A's marks were closer together and less spread out, showing more consistent performance.

🎯 Exam Tip: To compare consistency between different datasets, always use the coefficient of variation. A smaller C.V. value indicates greater consistency.

 

Question 7. If for distribution \( \Sigma(x – 7) = 3 \), \( \Sigma(x − 7)² = 57 \) and total number of item is 20. find the mean and standard deviation.
Answer:We are given: \( \Sigma(x - 7) = 3 \) \( \Sigma(x - 7)^2 = 57 \) Number of items \( n = 20 \) First, let's find the mean \( \bar{x} \). We know that \( \Sigma(x - 7) = \Sigma x - \Sigma 7 \).
\( \Sigma x - 7n = 3 \) Substitute \( n = 20 \):
\( \Sigma x - 7(20) = 3 \)
\( \Sigma x - 140 = 3 \)
\( \Sigma x = 140 + 3 \)
\( \Sigma x = 143 \) The arithmetic mean \( \bar{x} = \frac{\Sigma x}{n} \):
\( \bar{x} = \frac{143}{20} \)
\( \bar{x} = 7.15 \) This is the average value of all the data points in the distribution. Next, let's find the standard deviation \( \sigma \). We use the formula for standard deviation:
\[ \sigma = \sqrt{\frac{\Sigma x^2}{n} - \left(\frac{\Sigma x}{n}\right)^2} \] We need to find \( \Sigma x^2 \). We can use the given \( \Sigma(x - 7)^2 = 57 \).
\( \Sigma(x^2 - 14x + 49) = 57 \)
\( \Sigma x^2 - \Sigma 14x + \Sigma 49 = 57 \)
\( \Sigma x^2 - 14 \Sigma x + 49n = 57 \) Substitute \( \Sigma x = 143 \) and \( n = 20 \):
\( \Sigma x^2 - 14(143) + 49(20) = 57 \)
\( \Sigma x^2 - 2002 + 980 = 57 \)
\( \Sigma x^2 - 1022 = 57 \)
\( \Sigma x^2 = 57 + 1022 \)
\( \Sigma x^2 = 1079 \) Now, substitute \( \Sigma x^2 \) and \( \bar{x} \) into the standard deviation formula:
\[ \sigma = \sqrt{\frac{1079}{20} - (7.15)^2} \]
\[ \sigma = \sqrt{53.95 - 51.1225} \]
\[ \sigma = \sqrt{2.8275} \]
\[ \sigma \approx 1.6815 \]
\[ \sigma \approx 1.68 \] So, the arithmetic mean is 7.15 and the standard deviation is 1.68. Standard deviation tells us how much the data points typically differ from the mean.
In simple words: We first found the total sum of the numbers and then their average (mean). Then, we used another sum given in the problem to find out how spread out the numbers are, which is called the standard deviation.

🎯 Exam Tip: Remember that adding or subtracting a constant from each value in a dataset changes the mean but does not affect the standard deviation. This is useful for checking calculations.

 

Question 8. Two unbiased dice are rolled once. Find the probability of getting
(i) a sum 8
(ii) a doublet
(iii) a sum greater than 8
Answer:When two unbiased dice are rolled once, the total number of possible outcomes (sample space) is \( n(S) = 6 \times 6 = 36 \). Each roll is independent of the other, giving 36 unique pairs. (i) Let A be the event of getting a sum of 8. The outcomes that sum to 8 are: {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}. The number of outcomes for event A is \( n(A) = 5 \). The probability of event A is:
\[ P(A) = \frac{n(A)}{n(S)} = \frac{5}{36} \] (ii) Let B be the event of getting a doublet. A doublet means both dice show the same number. The outcomes for event B are: {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}. The number of outcomes for event B is \( n(B) = 6 \). The probability of event B is:
\[ P(B) = \frac{n(B)}{n(S)} = \frac{6}{36} = \frac{1}{6} \] (iii) Let C be the event of getting a sum greater than 8. This means the sum can be 9, 10, 11, or 12. The outcomes for event C are: For sum 9: {(3, 6), (4, 5), (5, 4), (6, 3)} For sum 10: {(4, 6), (5, 5), (6, 4)} For sum 11: {(5, 6), (6, 5)} For sum 12: {(6, 6)} So, C = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}. The number of outcomes for event C is \( n(C) = 10 \). The probability of event C is:
\[ P(C) = \frac{n(C)}{n(S)} = \frac{10}{36} = \frac{5}{18} \]
In simple words: When two dice are rolled, there are 36 possible results. We count how many of these results match what we want (like getting a sum of 8 or both dice showing the same number) and then divide that count by 36 to get the chance of it happening.

🎯 Exam Tip: Always list all possible outcomes for events clearly to avoid missing any pairs, especially for sums and doublets when rolling dice.

 

Question 9. A die is thrown twice. Find the probability that atleast one of the two throws conies up with the number 5 (use addition theorem).
Answer:When a die is thrown twice, the total number of possible outcomes (sample space) is \( n(S) = 6 \times 6 = 36 \). Let A be the event that 5 comes up on the first throw. A = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} The number of outcomes for event A is \( n(A) = 6 \). So, \( P(A) = \frac{n(A)}{n(S)} = \frac{6}{36} \). Let B be the event that 5 comes up on the second throw. B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)} The number of outcomes for event B is \( n(B) = 6 \). So, \( P(B) = \frac{n(B)}{n(S)} = \frac{6}{36} \). The event \( A \cap B \) means 5 comes up on both the first and second throws. \( A \cap B = \{(5, 5)\} \) The number of outcomes for \( A \cap B \) is \( n(A \cap B) = 1 \). So, \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{36} \). We need to find the probability that at least one of the two throws shows a 5, which is \( P(A \cup B) \). Using the Addition Theorem for probabilities:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = \frac{6}{36} + \frac{6}{36} - \frac{1}{36} \]
\[ P(A \cup B) = \frac{6 + 6 - 1}{36} \]
\[ P(A \cup B) = \frac{11}{36} \]
In simple words: We wanted to find the chance of rolling a 5 at least once with two dice. We found the chance of getting a 5 on the first roll, the chance of getting a 5 on the second roll, and the chance of getting a 5 on both. Then, we added the first two chances and subtracted the chance of getting 5 on both to avoid counting it twice.

🎯 Exam Tip: When using the addition theorem for "at least one" events, always remember to subtract the probability of the intersection to avoid double-counting outcomes that satisfy both events.

 

Question 10. Let A, B, C be any three mutually exclusive and exhaustive events such that \( P(B) = \frac{3}{2} P(A) \) and \( P(C) = \frac{1}{2} P(B) \). Find P(A).
Answer:Let \( P(A) = p \). Given that \( P(B) = \frac{3}{2} P(A) \). Substitute \( P(A) = p \):
\( P(B) = \frac{3}{2} p \) Given that \( P(C) = \frac{1}{2} P(B) \). Substitute \( P(B) = \frac{3}{2} p \):
\( P(C) = \frac{1}{2} \left(\frac{3}{2} p\right) \)
\( P(C) = \frac{3}{4} p \) Since A, B, and C are mutually exclusive and exhaustive events, their probabilities must add up to 1 (because they cover all possible outcomes). This means that if these events are the only possibilities, one of them must happen.
\( P(A) + P(B) + P(C) = 1 \) Substitute the expressions for \( P(A) \), \( P(B) \), and \( P(C) \) in terms of \( p \):
\( p + \frac{3}{2} p + \frac{3}{4} p = 1 \) To solve for \( p \), find a common denominator, which is 4:
\( \frac{4p}{4} + \frac{6p}{4} + \frac{3p}{4} = 1 \)
\( \frac{4p + 6p + 3p}{4} = 1 \)
\( \frac{13p}{4} = 1 \) Multiply both sides by 4:
\( 13p = 4 \) Divide by 13:
\( p = \frac{4}{13} \) Therefore, \( P(A) = \frac{4}{13} \).
In simple words: We used the given relationships between the probabilities of three events (A, B, C) and the fact that they must add up to 1 to find the probability of event A. This is like solving for an unknown part when you know how the parts relate to each other and what the total should be.

🎯 Exam Tip: For mutually exclusive and exhaustive events, remember that the sum of their individual probabilities must always equal 1. This is a key principle in probability.

 

Question 11. A bag contains 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rusted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.
Answer:First, let's find the total number of items in the bag: Total items \( n(S) = 50 \text{ bolts} + 150 \text{ nuts} = 200 \). Next, let's determine the number of rusted items and bolts: Number of rusted bolts = \( \frac{1}{2} \times 50 = 25 \). Number of rusted nuts = \( \frac{1}{2} \times 150 = 75 \). Let A be the event that the chosen item is rusted. The number of rusted items \( n(A) = \text{rusted bolts} + \text{rusted nuts} = 25 + 75 = 100 \). The probability of choosing a rusted item is:
\[ P(A) = \frac{n(A)}{n(S)} = \frac{100}{200} \] Let B be the event that the chosen item is a bolt. The number of bolts \( n(B) = 50 \). The probability of choosing a bolt is:
\[ P(B) = \frac{n(B)}{n(S)} = \frac{50}{200} \] The event \( A \cap B \) is choosing an item that is both rusted AND a bolt (i.e., a rusted bolt). The number of rusted bolts \( n(A \cap B) = 25 \). The probability of choosing a rusted bolt is:
\[ P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{25}{200} \] We need to find the probability that the item is rusted OR a bolt, which is \( P(A \cup B) \). Using the Addition Theorem for probabilities:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = \frac{100}{200} + \frac{50}{200} - \frac{25}{200} \] Combine the fractions:
\[ P(A \cup B) = \frac{100 + 50 - 25}{200} \]
\[ P(A \cup B) = \frac{150 - 25}{200} \]
\[ P(A \cup B) = \frac{125}{200} \] Simplify the fraction by dividing the numerator and denominator by 25:
\[ P(A \cup B) = \frac{5}{8} \]
In simple words: We first found the total number of items. Then, we calculated the chance of picking a rusted item and the chance of picking a bolt. Since some bolts are also rusted, we subtracted the chance of picking a rusted bolt so we don't count them twice.

🎯 Exam Tip: Clearly define your events A and B, and their intersection \( A \cap B \), before applying the addition rule. This helps avoid confusion and ensures correct calculation, especially in "or" scenarios.

 

Question 12. Two dice are rolled simultaneously. Find the probability that that sum of the numbers on the faces is neither divisible by 3 nor by 4.
Answer:When two dice are rolled simultaneously, the total number of possible outcomes (sample space) is \( n(S) = 6 \times 6 = 36 \). Let A be the event that the sum of the numbers is divisible by 3. The possible sums divisible by 3 are 3, 6, 9, 12. The outcomes for A are: Sum 3: {(1, 2), (2, 1)} Sum 6: {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)} Sum 9: {(3, 6), (6, 3), (4, 5), (5, 4)} Sum 12: {(6, 6)} So, A = {(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (3, 6), (6, 3), (4, 5), (5, 4), (6, 6)}. The number of outcomes for event A is \( n(A) = 12 \). The probability of event A is \( P(A) = \frac{12}{36} \). Let B be the event that the sum of the numbers is divisible by 4. The possible sums divisible by 4 are 4, 8, 12. The outcomes for B are: Sum 4: {(1, 3), (2, 2), (3, 1)} Sum 8: {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} Sum 12: {(6, 6)} So, B = {(1, 3), (2, 2), (3, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}. The number of outcomes for event B is \( n(B) = 9 \). The probability of event B is \( P(B) = \frac{9}{36} \). The event \( A \cap B \) is when the sum is divisible by both 3 and 4 (i.e., divisible by 12). The only outcome for \( A \cap B \) is: {(6, 6)}. The number of outcomes for \( A \cap B \) is \( n(A \cap B) = 1 \). The probability of event \( A \cap B \) is \( P(A \cap B) = \frac{1}{36} \). Now, find the probability that the sum is divisible by 3 OR 4, which is \( P(A \cup B) \). Using the Addition Theorem:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = \frac{12}{36} + \frac{9}{36} - \frac{1}{36} \]
\[ P(A \cup B) = \frac{12 + 9 - 1}{36} \]
\[ P(A \cup B) = \frac{20}{36} \] Finally, we need the probability that the sum is NEITHER divisible by 3 NOR by 4. This is the complement of \( P(A \cup B) \):
\[ P(\text{neither A nor B}) = 1 - P(A \cup B) \]
\[ P(\text{neither A nor B}) = 1 - \frac{20}{36} \]
\[ P(\text{neither A nor B}) = \frac{36}{36} - \frac{20}{36} \]
\[ P(\text{neither A nor B}) = \frac{16}{36} \] Simplify the fraction by dividing numerator and denominator by 4:
\[ P(\text{neither A nor B}) = \frac{4}{9} \]
In simple words: We listed all ways to get a sum divisible by 3 and all ways to get a sum divisible by 4. We found the chance of either of these happening. Then, to find the chance of NEITHER happening, we subtracted that probability from 1 (which represents 100% certainty).

🎯 Exam Tip: When a question asks for "neither A nor B", it usually implies calculating \( 1 - P(A \cup B) \). Make sure to correctly identify the intersection of events for the addition theorem.

 

Question 13. In a class, 40% of the students participated in Mathematics-quiz, 30% in Science - quiz and 10% in both the quiz programmes. If a students is selected at random from the class, find the probability that the students participated in Mathematics or science or both quiz programmes.
Answer:Let's assume the total number of students in the class (sample space) is \( n(S) = 100 \). Let A be the event that a student participated in the Mathematics-quiz. Given that 40% participated in Math-quiz, so \( n(A) = 40 \). The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{40}{100} \). Let B be the event that a student participated in the Science-quiz. Given that 30% participated in Science-quiz, so \( n(B) = 30 \). The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{30}{100} \). Let \( A \cap B \) be the event that a student participated in both quizzes. Given that 10% participated in both, so \( n(A \cap B) = 10 \). The probability of event \( A \cap B \) is \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{10}{100} \). We need to find the probability that a student participated in Mathematics OR Science OR both. This is represented by \( P(A \cup B) \). Using the Addition Theorem for probabilities:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = \frac{40}{100} + \frac{30}{100} - \frac{10}{100} \] Combine the fractions:
\[ P(A \cup B) = \frac{40 + 30 - 10}{100} \]
\[ P(A \cup B) = \frac{70 - 10}{100} \]
\[ P(A \cup B) = \frac{60}{100} \] Simplify the fraction:
\[ P(A \cup B) = \frac{3}{5} \]
In simple words: We found the chance that a student did the Math quiz, the Science quiz, or both. We added the chances for each quiz and then subtracted the chance of doing both so we don't count students who did both twice.

🎯 Exam Tip: When given percentages, treat the total as 100 to easily convert them into probabilities. Always subtract the intersection when finding the probability of "A or B" to avoid overcounting.

 

Question 14. A two digit number is formed with the digits 2, 5, 9 (repetition is allowed). Find the probability that the number is divisible by 2 or 5.
Answer:First, let's list all possible two-digit numbers formed using the digits 2, 5, 9, with repetition allowed. The first digit can be 2, 5, or 9 (3 choices). The second digit can be 2, 5, or 9 (3 choices). So, the total number of possible two-digit numbers (sample space) is \( n(S) = 3 \times 3 = 9 \). The numbers are: {22, 25, 29, 52, 55, 59, 92, 95, 99}. Let A be the event that the number is divisible by 2 (i.e., an even number). Numbers ending in 2 are divisible by 2. A = {22, 52, 92}. The number of outcomes for event A is \( n(A) = 3 \). The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{3}{9} \). Let B be the event that the number is divisible by 5 (i.e., ends in 5). Numbers ending in 5 are divisible by 5. B = {25, 55, 95}. The number of outcomes for event B is \( n(B) = 3 \). The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{3}{9} \). Now, let's find the intersection \( A \cap B \), which means the number is divisible by both 2 and 5 (i.e., divisible by 10). For a number to be divisible by 10, it must end in 0. None of our possible two-digit numbers end in 0. So, \( A \cap B = \{\} \) (an empty set). The number of outcomes for \( A \cap B \) is \( n(A \cap B) = 0 \). The probability of event \( A \cap B \) is \( P(A \cap B) = \frac{0}{9} = 0 \). Since \( P(A \cap B) = 0 \), events A and B are mutually exclusive. We need to find the probability that the number is divisible by 2 OR 5, which is \( P(A \cup B) \). Using the Addition Theorem for mutually exclusive events:
\[ P(A \cup B) = P(A) + P(B) \]
\[ P(A \cup B) = \frac{3}{9} + \frac{3}{9} \]
\[ P(A \cup B) = \frac{6}{9} \] Simplify the fraction by dividing the numerator and denominator by 3:
\[ P(A \cup B) = \frac{2}{3} \]
In simple words: We listed all the two-digit numbers we could make. Then, we found the chance of picking a number that can be divided by 2, and the chance of picking a number that can be divided by 5. Since no number could be divided by both, we just added these two chances together.

🎯 Exam Tip: When repetition is allowed, remember that each position in the number has the full set of digit choices. Carefully list all possible outcomes for small sample spaces to ensure accuracy.

 

Question 15. The probability that A, B and C can solve a problem are \( \frac{4}{5}, \frac{2}{3} \) and \( \frac{3}{7} \) respectively. The probability of the problem being solved by A and B is \( \frac{8}{15} \), B and C is \( \frac{2}{7} \), A and C is \( \frac{12}{35} \). The probability of the problem being solved by all the three is \( \frac{8}{35} \). Find the probability that the problem can be solved by atleast one of them.
Answer:We are given the following probabilities:
\( P(A) = \frac{4}{5} \) (Probability A solves the problem)
\( P(B) = \frac{2}{3} \) (Probability B solves the problem)
\( P(C) = \frac{3}{7} \) (Probability C solves the problem)
\( P(A \cap B) = \frac{8}{15} \) (Probability A and B both solve it)
\( P(B \cap C) = \frac{2}{7} \) (Probability B and C both solve it)
\( P(A \cap C) = \frac{12}{35} \) (Probability A and C both solve it)
\( P(A \cap B \cap C) = \frac{8}{35} \) (Probability A, B, and C all solve it) We need to find the probability that the problem can be solved by at least one of them, which is \( P(A \cup B \cup C) \). We use the general Addition Theorem for three events:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C) \] Now, substitute the given probabilities into the formula:
\[ P(A \cup B \cup C) = \frac{4}{5} + \frac{2}{3} + \frac{3}{7} - \frac{8}{15} - \frac{2}{7} - \frac{12}{35} + \frac{8}{35} \] To add and subtract these fractions, we need a common denominator. The least common multiple (LCM) of 5, 3, 7, 15, and 35 is 105. Convert each fraction to have a denominator of 105:
\( \frac{4}{5} = \frac{4 \times 21}{5 \times 21} = \frac{84}{105} \)
\( \frac{2}{3} = \frac{2 \times 35}{3 \times 35} = \frac{70}{105} \)
\( \frac{3}{7} = \frac{3 \times 15}{7 \times 15} = \frac{45}{105} \)
\( \frac{8}{15} = \frac{8 \times 7}{15 \times 7} = \frac{56}{105} \)
\( \frac{2}{7} = \frac{2 \times 15}{7 \times 15} = \frac{30}{105} \)
\( \frac{12}{35} = \frac{12 \times 3}{35 \times 3} = \frac{36}{105} \)
\( \frac{8}{35} = \frac{8 \times 3}{35 \times 3} = \frac{24}{105} \) Now, substitute these equivalent fractions back into the Addition Theorem formula:
\[ P(A \cup B \cup C) = \frac{84}{105} + \frac{70}{105} + \frac{45}{105} - \frac{56}{105} - \frac{30}{105} - \frac{36}{105} + \frac{24}{105} \] Combine the numerators:
\[ P(A \cup B \cup C) = \frac{84 + 70 + 45 - 56 - 30 - 36 + 24}{105} \] First, sum the positive terms: \( 84 + 70 + 45 + 24 = 223 \). Then, sum the negative terms: \( 56 + 30 + 36 = 122 \).
\[ P(A \cup B \cup C) = \frac{223 - 122}{105} \]
\[ P(A \cup B \cup C) = \frac{101}{105} \]
In simple words: We used a special formula to find the chance that at least one of the three people could solve the problem. This formula adds up each person's chance, subtracts the chances of any two solving it together (to avoid double-counting), and then adds back the chance of all three solving it (because that was subtracted too many times).

🎯 Exam Tip: For problems involving "at least one" of multiple events, the inclusion-exclusion principle (generalized addition theorem) is essential. Carefully list all individual, pairwise, and triple intersections.