Samacheer Kalvi Class 10 Maths Solutions Chapter 8 Statistics and Probability Exercise 8

Question 1. The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies \( f_1 \) and \( f_2 \).
Answer:The sum of all frequencies is \( \Sigma f_i = 50 \). Let the missing frequencies be \( f_1 \) and \( f_2 \). First, we construct a table to find the mid-values and calculate \( f_i x_i \):

From the total frequency given: \( 30 + f_1 + f_2 = 50 \)
\( \implies f_1 + f_2 = 50 - 30 \)
\( \implies f_1 + f_2 = 20 \) ...(1)

The mean \( \bar{x} \) is given as 62.8. The formula for the mean is \( \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} \). \( 62.8 = \frac{2060 + 30f_1 + 70f_2}{50} \) Now, we multiply both sides by 50 to clear the denominator: \( 62.8 \times 50 = 2060 + 30f_1 + 70f_2 \) \( 3140 = 2060 + 30f_1 + 70f_2 \) Rearrange the terms to get an equation in \( f_1 \) and \( f_2 \): \( 30f_1 + 70f_2 = 3140 - 2060 \)
\( \implies 30f_1 + 70f_2 = 1080 \) Divide the entire equation by 10 to simplify:
\( \implies 3f_1 + 7f_2 = 108 \) ...(2)

Now we have a system of two linear equations:

(1) \( f_1 + f_2 = 20 \) (2) \( 3f_1 + 7f_2 = 108 \)

To solve this, multiply equation (1) by 3: \( 3 \times (f_1 + f_2) = 3 \times 20 \)
\( \implies 3f_1 + 3f_2 = 60 \) ...(3)

Subtract equation (3) from equation (2): \( (3f_1 + 7f_2) - (3f_1 + 3f_2) = 108 - 60 \) \( 4f_2 = 48 \)
\( \implies f_2 = \frac{48}{4} \)
\( \implies f_2 = 12 \)

Substitute the value of \( f_2 = 12 \) back into equation (1): \( f_1 + 12 = 20 \)
\( \implies f_1 = 20 - 12 \)
\( \implies f_1 = 8 \)

Thus, the missing frequencies are \( f_1 = 8 \) and \( f_2 = 12 \). These values help complete the frequency distribution accurately, maintaining the given mean and total frequency.
In simple words: We used the total number of frequencies and the average (mean) of the distribution to set up two math problems. By solving these, we found the two missing numbers in the frequency list.

🎯 Exam Tip: When given mean and total frequency with missing values, always form two simultaneous equations (one from total frequency, one from the mean formula) and solve them.

Question 2. The diameter of circles (in mm) drawn in the design are given below. Calculate the standard deviation.
Answer:The assumed mean \( A \) is given as 42.5. We use this to calculate the deviations \( d_i \) and then the standard deviation. The standard deviation measures how spread out the data is from its average value.

The formula for standard deviation \( \sigma \) using the assumed mean method is: \( \sigma = \sqrt{\frac{\Sigma f_i d_i^2}{\Sigma f_i} - \left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2} \) Now, we substitute the values from the table into the formula: \( \sigma = \sqrt{\frac{3184}{100} - \left(\frac{100}{100}\right)^2} \) First, calculate the terms inside the square root: \( \sigma = \sqrt{31.84 - (1)^2} \) \( \sigma = \sqrt{31.84 - 1} \) \( \sigma = \sqrt{30.84} \) Finally, calculate the square root: \( \sigma \approx 5.5533 \) Rounding to two decimal places, the standard deviation \( \sigma = 5.55 \). This indicates the average distance of each data point from the mean.
In simple words: We found out how much the circle diameters spread out from their average size. We used a special math formula and a table to do this, finding that the diameters vary by about 5.55 mm on average.

🎯 Exam Tip: Always double-check your calculations for \( d_i \) and \( f_i d_i^2 \) to avoid errors. Ensure the assumed mean is correctly subtracted for all \( x_i \) values.

Question 3. The frequency distribution is given below. In table k is a positive integer, has a variance of 160. Determine the value of k.
Answer:We are given the frequency distribution and that the variance is 160. We need to find the value of \( k \). We will use the assumed mean method for variance. From the table, it looks like the assumed mean is \( 3k \), so \( d = x - 3k \).

The formula for variance \( \sigma^2 \) using the assumed mean method is: \( \sigma^2 = \frac{\Sigma f_i d_i^2}{\Sigma f_i} - \left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2 \) Given \( \sigma^2 = 160 \), and from the table, \( \Sigma f_i = 7 \), \( \Sigma f_i d_i = k \), \( \Sigma f_i d_i^2 = 23k^2 \). Substitute these values into the variance formula: \( 160 = \frac{23k^2}{7} - \left(\frac{k}{7}\right)^2 \)
\( \implies 160 = \frac{23k^2}{7} - \frac{k^2}{49} \) To combine the fractions, we find a common denominator, which is 49: \( 160 = \frac{23k^2 \times 7}{49} - \frac{k^2}{49} \)
\( \implies 160 = \frac{161k^2 - k^2}{49} \)
\( \implies 160 = \frac{160k^2}{49} \) Now, we can solve for \( k^2 \): \( 160 \times 49 = 160k^2 \)
\( \implies k^2 = \frac{160 \times 49}{160} \)
\( \implies k^2 = 49 \) Since \( k \) is a positive integer, we take the positive square root: \( k = \sqrt{49} \)
\( \implies k = 7 \)

The value of \( k \) is 7. This result shows how the variance relates to the structure of the frequency distribution defined by \( k \).
In simple words: We used a special math formula for how much numbers spread out (variance), along with the numbers from the table. By putting all the given information into the formula, we were able to find the value of \( k \).

🎯 Exam Tip: Be careful with algebraic manipulation, especially when dealing with fractions and squaring terms. Ensure you simplify fractions correctly before solving for the unknown variable.

Question 4. The standard deviation of some temperature data in degree Celsius (°C) is 5. If the data were converted into degree Fahrenheit (°F) then what is the variance?
Answer:Given: Standard deviation of temperature in Celsius, \( \sigma_C = 5 \). The formula to convert Celsius to Fahrenheit is \( F = C \times 1.8 + 32 \). When we transform data, adding a constant (like +32) does not change the standard deviation. However, multiplying by a constant (like 1.8) multiplies the standard deviation by that constant.

So, the standard deviation in Fahrenheit (\( \sigma_F \)) is: \( \sigma_F = 1.8 \times \sigma_C \) \( \sigma_F = 1.8 \times 5 \)
\( \implies \sigma_F = 9 \)

Variance is the square of the standard deviation. Variance in Fahrenheit \( = \sigma_F^2 \) Variance \( = 9^2 \)
\( \implies \) Variance \( = 81 \)

Therefore, the variance of the temperature data in Fahrenheit is 81. This highlights how scale changes (multiplication) impact variability measures, while shifts (addition) do not affect standard deviation.
In simple words: If you change temperature from Celsius to Fahrenheit, you multiply it by 1.8 and add 32. The "add 32" part does not change how spread out the temperatures are. But multiplying by 1.8 means the spread also gets multiplied by 1.8. So, the new spread (standard deviation) is 9. The variance is simply this spread number multiplied by itself.

🎯 Exam Tip: Remember that adding or subtracting a constant to all data points does not change the standard deviation, but multiplying or dividing by a constant scales the standard deviation by the same factor.

Question 5. If for a distribution, \( \Sigma (x - 5) = 3 \), \( \Sigma (x - 5)^2 = 43 \), and total number of observations is 18, find the mean and standard deviation.
Answer:Given: \( \Sigma (x - 5) = 3 \) \( \Sigma (x - 5)^2 = 43 \) Number of observations \( n = 18 \)

**Part 1: Find the mean (\( \bar{x} \))** We expand the first given sum: \( \Sigma (x - 5) = 3 \)
\( \implies \Sigma x - \Sigma 5 = 3 \) Since \( \Sigma 5 \) means adding 5 for each of the \( n \) observations, \( \Sigma 5 = 5 \times n \). \( \Sigma x - (5 \times 18) = 3 \)
\( \implies \Sigma x - 90 = 3 \)
\( \implies \Sigma x = 3 + 90 \)
\( \implies \Sigma x = 93 \) Now, calculate the mean \( \bar{x} \): \( \bar{x} = \frac{\Sigma x}{n} = \frac{93}{18} \)
\( \implies \bar{x} = 5.166... \) Rounding to two decimal places, the arithmetic mean \( \bar{x} = 5.17 \).

**Part 2: Find the standard deviation (\( \sigma \))** The formula for standard deviation when deviations from an arbitrary constant 'a' are given is: \( \sigma = \sqrt{\frac{\Sigma (x - a)^2}{n} - \left(\frac{\Sigma (x - a)}{n}\right)^2} \) Here, the constant \( a = 5 \). Substitute the given values: \( \sigma = \sqrt{\frac{43}{18} - \left(\frac{3}{18}\right)^2} \) First, we calculate the values: \( \frac{43}{18} \approx 2.3888... \) \( \left(\frac{3}{18}\right)^2 = \left(\frac{1}{6}\right)^2 = \frac{1}{36} \approx 0.0277... \) So, \( \sigma = \sqrt{2.3888... - 0.0277...} \) \( \sigma = \sqrt{2.3611...} \)
\( \implies \sigma \approx 1.5366 \) Rounding to two decimal places, the standard deviation \( \sigma = 1.54 \). (Note: if using the source's intermediate calculation \( \sqrt{29.06 - 26.73} = \sqrt{2.33} \), then \( \sigma \approx 1.53 \). We will follow the source's final rounded answer.)

Alternatively, using the other formula for standard deviation: \( \sigma = \sqrt{\frac{\Sigma x^2}{n} - \left(\frac{\Sigma x}{n}\right)^2} \) We first find \( \Sigma x^2 \): \( \Sigma (x - 5)^2 = 43 \)
\( \implies \Sigma (x^2 - 10x + 25) = 43 \)
\( \implies \Sigma x^2 - 10 \Sigma x + \Sigma 25 = 43 \) We know \( \Sigma x = 93 \) and \( \Sigma 25 = 25 \times 18 = 450 \). \( \Sigma x^2 - 10(93) + 450 = 43 \)
\( \implies \Sigma x^2 - 930 + 450 = 43 \)
\( \implies \Sigma x^2 - 480 = 43 \)
\( \implies \Sigma x^2 = 43 + 480 \)
\( \implies \Sigma x^2 = 523 \) Now, substitute \( \Sigma x^2 \) and \( \Sigma x \) into the standard deviation formula: \( \sigma = \sqrt{\frac{523}{18} - \left(\frac{93}{18}\right)^2} \) \( \sigma = \sqrt{29.055... - (5.166...)^2} \) \( \sigma = \sqrt{29.06 - 26.73} \) (using rounded values as in the source) \( \sigma = \sqrt{2.33} \) \( \sigma \approx 1.526 \) Rounding to two decimal places, the standard deviation \( \sigma = 1.53 \). The mean is 5.17 and the standard deviation is 1.53. These values help understand the central tendency and spread of the distribution.
In simple words: We used the given sums and the total number of items to first find the average (mean) of the numbers. Then, using another formula with the given sums, we figured out how much the numbers spread out from that average (standard deviation).

🎯 Exam Tip: When given sums like \( \Sigma(x-a) \) and \( \Sigma(x-a)^2 \), use the appropriate formula for standard deviation which directly incorporates these sums to simplify calculations.

Question 6. Prices in various places of two cities are given below. In which city, prices were more stable?
Answer:To find which city has more stable prices, we need to calculate the coefficient of variation (CV) for both cities. A lower coefficient of variation indicates greater stability in prices.

**For City A:** Prices (in ascending order): 16, 19, 20, 22, 23. Number of observations \( n = 5 \). We use an assumed mean of 20 (as per the source solution, which happens to be the actual mean).
Mean \( \bar{x} = \frac{16+19+20+22+23}{5} = \frac{100}{5} = 20 \).

Standard deviation \( \sigma_A = \sqrt{\frac{\Sigma d_i^2}{n} - \left(\frac{\Sigma d_i}{n}\right)^2} \) \( \sigma_A = \sqrt{\frac{30}{5} - \left(\frac{0}{5}\right)^2} \) \( \sigma_A = \sqrt{6 - 0} = \sqrt{6} \) \( \sigma_A \approx 2.45 \) Coefficient of Variation for City A: \( CV_A = \frac{\sigma_A}{\bar{x}} \times 100 = \frac{2.45}{20} \times 100 = 0.1225 \times 100 = 12.25\% \)

**For City B:** Prices (in ascending order): 10, 12, 15, 18, 20. Number of observations \( n = 5 \). We use an assumed mean of 15 (as per the source solution, which happens to be the actual mean).
Mean \( \bar{x} = \frac{10+12+15+18+20}{5} = \frac{75}{5} = 15 \).

Standard deviation \( \sigma_B = \sqrt{\frac{\Sigma d_i^2}{n} - \left(\frac{\Sigma d_i}{n}\right)^2} \) \( \sigma_B = \sqrt{\frac{68}{5} - \left(\frac{0}{5}\right)^2} \) \( \sigma_B = \sqrt{13.6 - 0} = \sqrt{13.6} \) \( \sigma_B \approx 3.69 \) Coefficient of Variation for City B: \( CV_B = \frac{\sigma_B}{\bar{x}} \times 100 = \frac{3.69}{15} \times 100 = 0.246 \times 100 = 24.6\% \)

**Comparison:** \( CV_A = 12.25\% \) \( CV_B = 24.6\% \) Since \( CV_A < CV_B \) (12.25% is less than 24.6%), the prices in City A are more stable than in City B. The coefficient of variation helps to compare the relative variability between different data sets.
In simple words: We calculated a special number called the "coefficient of variation" for both cities, which tells us how much the prices change compared to their average. City A had a smaller number, meaning its prices were more steady and less all over the place.

🎯 Exam Tip: Remember that a lower coefficient of variation (CV) indicates more consistency or stability, while a higher CV indicates greater variability. Ensure to calculate both the mean and standard deviation correctly for each dataset before computing the CV.

Question 7. If the range and coefficient of range of the data are 20 and 0.2 respectively, then find the largest and smallest values of the data.
Answer:Given: Range \( R = 20 \) Coefficient of Range \( C_R = 0.2 \) Let \( L \) be the largest value and \( S \) be the smallest value of the data.

The formula for Range is: \( R = L - S \) So, we have: \( L - S = 20 \) ...(1)

The formula for Coefficient of Range is: \( C_R = \frac{L - S}{L + S} \) Substitute the given values into this formula: \( 0.2 = \frac{20}{L + S} \) Now, we can solve for \( L + S \): \( L + S = \frac{20}{0.2} \) To simplify the fraction, multiply the numerator and denominator by 10: \( L + S = \frac{200}{2} \)
\( \implies L + S = 100 \) ...(2)

Now we have a system of two linear equations:

(1) \( L - S = 20 \) (2) \( L + S = 100 \)

Add equation (1) and equation (2): \( (L - S) + (L + S) = 20 + 100 \) \( 2L = 120 \)
\( \implies L = \frac{120}{2} \)
\( \implies L = 60 \)

Substitute the value of \( L = 60 \) back into equation (2): \( 60 + S = 100 \)
\( \implies S = 100 - 60 \)
\( \implies S = 40 \)

Therefore, the largest value of the data is 60 and the smallest value is 40. These two measures, range and coefficient of range, help describe the spread of data in different contexts.
In simple words: We used two clues: how far apart the highest and lowest numbers were (the range), and how this difference compared to their total (coefficient of range). By solving these two number puzzles together, we found out what the highest and lowest numbers were.

🎯 Exam Tip: When given both the range and coefficient of range, always set up two simultaneous equations using the formulas \( L-S \) and \( \frac{L-S}{L+S} \) to solve for the largest (L) and smallest (S) values.

Question 8. If two dice are rolled, then find the probability of getting the product of face value 6 or the difference of face values 5.
Answer:When two dice are rolled, the total number of possible outcomes is \( 6 \times 6 = 36 \). This is our sample space, \( n(S) = 36 \). The sample space consists of pairs like \((1,1), (1,2), \dots, (6,6)\).

Let A be the event of getting the product of face values as 6. The pairs whose product is 6 are: \( A = \{(1, 6), (2, 3), (3, 2), (6, 1)\} \) So, the number of outcomes in event A is \( n(A) = 4 \). The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{4}{36} \).

Let B be the event of getting the difference of face values as 5. The pairs whose difference is 5 are: \( B = \{(1, 6), (6, 1)\} \) So, the number of outcomes in event B is \( n(B) = 2 \). (Note: The source text shows \( n(B)=1 \) and \( B = \{(6,1)\} \). We will proceed with the logically correct \( n(B)=2 \) to avoid misinterpretation, as both (1,6) and (6,1) have a difference of 5). The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{2}{36} \).

Now, we need to find the outcomes common to both A and B (the intersection \( A \cap B \)): \( A \cap B = \{(1, 6), (6, 1)\} \) So, the number of outcomes in \( A \cap B \) is \( n(A \cap B) = 2 \). The probability of \( A \cap B \) is \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{36} \).

We want the probability of getting the product of face value 6 OR the difference of face values 5, which is \( P(A \cup B) \). We use the formula: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) \( P(A \cup B) = \frac{4}{36} + \frac{2}{36} - \frac{2}{36} \) \( P(A \cup B) = \frac{4 + 2 - 2}{36} \) \( P(A \cup B) = \frac{4}{36} \)
\( \implies P(A \cup B) = \frac{1}{9} \)

The probability of getting a product of 6 or a difference of 5 is \( \frac{1}{9} \). This calculation demonstrates how to combine probabilities of overlapping events using the inclusion-exclusion principle.
In simple words: When rolling two dice, there are 36 possible outcomes. We found all pairs that multiply to 6 and all pairs that have a difference of 5. We then counted the unique pairs from both lists. The chance of either of these things happening is 4 out of 36, which simplifies to 1 out of 9.

🎯 Exam Tip: For "OR" probability questions, always identify overlapping outcomes (intersection) and use the formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) to avoid double-counting.

Question 9. In a two children family, find the probability that there is at least one girl in a family.
Answer:For a family with two children, each child can be either a Boy (B) or a Girl (G). The sample space (S) lists all possible combinations: \( S = \{(Boy, Boy), (Boy, Girl), (Girl, Boy), (Girl, Girl)\} \) The total number of outcomes in the sample space is \( n(S) = 4 \).

Let A be the event that there is at least one girl in the family. "At least one girl" means one girl or two girls. The outcomes for this event are: \( A = \{(Boy, Girl), (Girl, Boy), (Girl, Girl)\} \) The number of outcomes in event A is \( n(A) = 3 \).

The probability of event A is calculated as: \( P(A) = \frac{n(A)}{n(S)} = \frac{3}{4} \)

Therefore, the probability of having at least one girl in a two-children family is \( \frac{3}{4} \). This simple example illustrates basic probability principles using a defined sample space.
In simple words: For two children, there are four possible combinations: boy-boy, boy-girl, girl-boy, or girl-girl. Out of these, three combinations have at least one girl. So, the chance of having at least one girl is 3 out of 4.

🎯 Exam Tip: Always list all possible outcomes (sample space) clearly. For "at least one" probabilities, it's often easier to count the desired outcomes directly or use the complement rule (1 - P(none)).

Question 10. A bag contains 5 white and some black balls. If the probability of drawing a black ball from the bag is twice the probability of drawing a white ball then find the number of black balls.
Answer:Let the number of black balls in the bag be \( x \). The number of white balls in the bag is 5. The total number of balls in the bag (sample space, \( n(S) \)) is the sum of black and white balls: \( n(S) = x + 5 \)

Let A be the event of drawing a black ball. The number of black balls is \( n(A) = x \). The probability of drawing a black ball is \( P(A) = \frac{n(A)}{n(S)} = \frac{x}{x + 5} \).

Let B be the event of drawing a white ball. The number of white balls is \( n(B) = 5 \). The probability of drawing a white ball is \( P(B) = \frac{n(B)}{n(S)} = \frac{5}{x + 5} \).

According to the problem statement, the probability of drawing a black ball is twice the probability of drawing a white ball: \( P(A) = 2 \times P(B) \) Substitute the probability expressions: \( \frac{x}{x + 5} = 2 \times \frac{5}{x + 5} \)
\( \implies \frac{x}{x + 5} = \frac{10}{x + 5} \) Since the denominators are equal and cannot be zero (as there are balls in the bag), we can equate the numerators: \( x = 10 \)

Alternatively, following a quadratic approach: If we multiply both sides by \( (x+5) \), this should simplify to \( x=10 \). However, if we were to multiply by \( (x+5)^2 \) and assume a quadratic form (as in some solutions to demonstrate the invalidity of a negative root for context-specific problems), it would be: \( x(x+5) = 10(x+5) \) \( x^2 + 5x = 10x + 50 \) Move all terms to one side to form a quadratic equation: \( x^2 + 5x - 10x - 50 = 0 \)
\( \implies x^2 - 5x - 50 = 0 \) Factorize the quadratic equation: \( (x - 10)(x + 5) = 0 \) This gives two possible values for \( x \): \( x - 10 = 0 \implies x = 10 \) \( x + 5 = 0 \implies x = -5 \) Since the number of balls cannot be negative, we discard \( x = -5 \). Therefore, the number of black balls is 10. This problem showcases how algebraic equations can be used to solve probability problems involving unknown quantities.
In simple words: We used math to figure out how many black balls were in the bag. We knew the chances of picking a black ball were double the chances of picking a white ball. By setting up a simple equation with this information, we found there were 10 black balls.

🎯 Exam Tip: Always define your variables clearly (like 'x' for black balls) and set up the probability ratios correctly. Remember that physical quantities like the number of balls cannot be negative, so discard any negative solutions.

Question 11. The probability that a student will pass the final examination in both English and Tamil is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Tamil examination?
Answer:Let A be the event that a student passes the English examination. Let B be the event that a student passes the Tamil examination.

Given probabilities: 1. Probability of passing both English and Tamil: \( P(A \cap B) = 0.5 \) 2. Probability of passing neither English nor Tamil: \( P(A' \cap B') = 0.1 \) 3. Probability of passing the English examination: \( P(A) = 0.75 \)

We need to find the probability of passing the Tamil examination, \( P(B) \).

From the probability of passing neither, we know that \( P(A' \cap B') = P((A \cup B)') \). So, \( P((A \cup B)') = 0.1 \) Using the complement rule, \( P(A \cup B) = 1 - P((A \cup B)') \). \( P(A \cup B) = 1 - 0.1 \)
\( \implies P(A \cup B) = 0.9 \)

Now, we use the formula for the probability of the union of two events: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) Substitute the known values into this formula: \( 0.9 = 0.75 + P(B) - 0.5 \) Combine the known probabilities on the right side: \( 0.9 = (0.75 - 0.5) + P(B) \) \( 0.9 = 0.25 + P(B) \) Now, solve for \( P(B) \): \( P(B) = 0.9 - 0.25 \)
\( \implies P(B) = 0.65 \)

Converting to fractions as in the source solution for demonstration: \( P(A \cap B) = \frac{1}{2} \) \( P(A) = \frac{3}{4} \) \( P(A \cup B) = 1 - \frac{1}{10} = \frac{9}{10} \) \( \frac{9}{10} = \frac{3}{4} + P(B) - \frac{1}{2} \) \( P(B) = \frac{9}{10} - \frac{3}{4} + \frac{1}{2} \) To add and subtract these fractions, find a common denominator, which is 20: \( P(B) = \frac{18}{20} - \frac{15}{20} + \frac{10}{20} \) \( P(B) = \frac{18 - 15 + 10}{20} \) \( P(B) = \frac{3 + 10}{20} \)
\( \implies P(B) = \frac{13}{20} \)

\( \frac{13}{20} = 0.65 \). Thus, the probability of passing the Tamil examination is 0.65 or \( \frac{13}{20} \). This problem effectively uses the principles of set theory and probability, including the complement rule and the union of events.
In simple words: We used clues about students passing English, Tamil, or neither. First, if 10% passed neither, then 90% passed at least one subject. Then, we used a formula that connects the chances of passing English, Tamil, and both. By putting in the known numbers, we found the chance of passing Tamil.

🎯 Exam Tip: When given "probability of passing neither," always use De Morgan's Law and the complement rule: \( P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) \). This is a common step to find \( P(A \cup B) \).

Question 12. The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting
(i) a diamond
(ii) a queen
(iii) a spade
(iv) a heart card bearing the number 5.
Answer:Initially, there are 52 cards in a standard deck. The King of spades, Queen of spades, and Jack of spades (3 cards) are removed. The remaining number of cards in the deck is \( 52 - 3 = 49 \). This new total becomes our sample space, so \( n(S) = 49 \).

**(i) Probability of getting a diamond:** There are 13 diamond cards in a standard deck. None of these were removed. Let A be the event of drawing a diamond card. \( n(A) = 13 \) The probability of drawing a diamond is \( P(A) = \frac{n(A)}{n(S)} = \frac{13}{49} \).

**(ii) Probability of getting a queen:** There are 4 queen cards in a standard deck (Queen of Hearts, Queen of Diamonds, Queen of Clubs, Queen of Spades). The Queen of spades was removed. So, the number of remaining queen cards is \( 4 - 1 = 3 \). Let B be the event of drawing a queen card. \( n(B) = 3 \) The probability of drawing a queen is \( P(B) = \frac{n(B)}{n(S)} = \frac{3}{49} \).

**(iii) Probability of getting a spade:** There are 13 spade cards in a standard deck. The King, Queen, and Jack of spades (3 cards) were removed. So, the number of remaining spade cards is \( 13 - 3 = 10 \). Let C be the event of drawing a spade card. \( n(C) = 10 \) The probability of drawing a spade is \( P(C) = \frac{n(C)}{n(S)} = \frac{10}{49} \).

**(iv) Probability of getting a heart card bearing the number 5:** There is only one '5 of Hearts' card in a standard deck. This card was not among the removed cards. Let D be the event of drawing a 5 of hearts card. \( n(D) = 1 \) The probability of drawing a 5 of hearts is \( P(D) = \frac{n(D)}{n(S)} = \frac{1}{49} \).

This problem illustrates how probabilities change when the total sample space and the number of favorable outcomes are modified.
In simple words: We started with a full deck of 52 cards but took out three specific spade cards. So, there were only 49 cards left. Then we calculated the chance of drawing different types of cards from this smaller deck. For example, the chance of drawing a diamond stayed the same because no diamonds were removed, but the chance of drawing a queen or a spade changed because some were removed.

🎯 Exam Tip: Always re-calculate the total number of outcomes (sample space, \( n(S) \)) and the number of favorable outcomes for each event after cards are removed or added to a deck.