Samacheer Kalvi Class 11 Maths Solutions Chapter 12 Introduction to Probability Theory Exercise 12.5

Question 1. Four persons are selected at random from a group of 3 men, 2 women and 4 children. The probability that exactly two of them are children is
(a) \( \frac{3}{4} \)
(b) \( \frac{10}{23} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{10}{21} \)
Answer: (d) \( \frac{10}{21} \)
In simple words: We need to find the chance that out of four people chosen from a mixed group (men, women, children), exactly two of them are children. This involves counting combinations of children and others, then dividing by the total possible combinations.

🎯 Exam Tip: Remember that combinations \( \text{nC}_r \) represent selecting 'r' items from 'n' without regard to order. The formula is \( \text{nC}_r = \frac{n!}{r!(n-r)!} \).

Question 2. A number is selected from the set {1, 2, 3, ....., 20}. The probability that the selected number is divisible by 3 or 4 is
(a) \( \frac{2}{5} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{2}{5} \)
(d) \( \frac{2}{5} \)
Answer: (c) \( \frac{2}{5} \)
In simple words: We are picking a number from 1 to 20. We want to find the chance that this number can be divided evenly by 3, or by 4, or by both. We count numbers divisible by 3, numbers divisible by 4, and then subtract numbers divisible by both to avoid double-counting.

🎯 Exam Tip: When finding the probability of 'A or B', use the formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Make sure to list all elements in the sample space and events A and B clearly.

Question 3. Three persons A, B, C fire a target simultaneously but independently. Their respective probabilities of hitting the target are \( \frac{3}{4} \), \( \frac{1}{2} \), \( \frac{5}{8} \). The probability that the target is hit by A or B but not by C is
(a) \( \frac{21}{64} \)
(b) \( \frac{7}{32} \)
(c) \( \frac{9}{64} \)
(d) \( \frac{7}{8} \)
Answer: (a) \( \frac{21}{64} \)
In simple words: Three people shoot at a target. We know how likely each person is to hit it. We need to find the chance that either person A or person B hits the target, but person C definitely misses. Since their shots are independent, we can multiply their individual probabilities after finding the probability of A or B hitting.

🎯 Exam Tip: For independent events, the probability of A and B happening is \( P(A \cap B) = P(A) \times P(B) \). The probability of an event not happening (complement) is \( P(A') = 1 - P(A) \).

Question 4. If A and B are any two events, then the probability that exactly one of them occur is
(a) \( P(A \cap B) + P(\overline{A} \cap \overline{B}) \)
(b) \( P(A \cap \overline{B}) + P(\overline{A} \cap B) \)
(c) \( P(A) + P(B) - P(A \cap B) \)
(d) \( P(A) + P(B) + 2P(A \cap B) \)
Answer: (b) \( P(A \cap \overline{B}) + P(\overline{A} \cap B) \)
In simple words: When we talk about "exactly one of them occurring," it means either event A happens and event B does not, OR event B happens and event A does not. This is like choosing one outcome from two possibilities.

🎯 Exam Tip: Visualizing with a Venn diagram helps for these types of questions. The regions for "exactly one of them occurs" are the parts of A and B that do not overlap.

Question 5. Let A and B be two events such that \( P(\overline{A \cup B}) = \frac{1}{6} \), \( P(A \cap B) = \frac{1}{4} \) and \( P(\overline{A}) = \frac{1}{4} \). Then the events A and B are
(a) Equally likely but not independent
(b) Independent but not equally likely
(c) Independent and equally likely
(d) Mutually inclusive and dependent
Answer: (b) Independent but not equally likely
In simple words: We are given some probabilities about two events, A and B. We need to figure out if these events are "equally likely" (meaning they have the same chance of happening) and if they are "independent" (meaning one happening doesn't affect the other). We use the given probabilities to find \( P(A) \) and \( P(B) \) and then check the conditions for independence and equal likelihood.

🎯 Exam Tip: Events A and B are independent if \( P(A \cap B) = P(A) \times P(B) \). They are equally likely if \( P(A) = P(B) \). Use the complement rule \( P(\overline{E}) = 1 - P(E) \) to find missing probabilities.

Question 6. Two items are chosen from a lot containing twelve items of which four are defective, then the probability that at least one of the item is defective
(a) \( \frac{19}{33} \)
(b) \( \frac{17}{33} \)
(c) \( \frac{23}{33} \)
(d) \( \frac{13}{33} \)
Answer: (a) \( \frac{19}{33} \)
In simple words: We pick two items from a group of twelve, where some are faulty. We want to find the chance that at least one of the two items we picked is faulty. This can be found by calculating the total ways to pick two items, and the ways to pick two items that are NOT faulty, then subtracting. Alternatively, we can calculate the probability of picking exactly one defective or exactly two defective items.

🎯 Exam Tip: The probability of "at least one" event can often be easily found by using the complement rule: \( P(\text{at least one defective}) = 1 - P(\text{no defective items}) \).

Question 7. A man has 3 fifty rupee notes, 4 hundred rupees notes and 6 five hundred rupees notes in his pocket. If 2 notes are taken at random, what are the odds in favour of both notes being of hundred rupee denomination?
(a) 1:12
(b) 12:1
(c) 13:1
(d) 1:3
Answer: (a) 1:12
In simple words: A man has different kinds of money notes. He pulls out two notes by chance. We want to know the chances that both the notes he picked are hundred rupee notes. "Odds in favour" means comparing the number of ways the event can happen to the number of ways it cannot happen.

🎯 Exam Tip: To calculate "odds in favour" for an event E, use the ratio \( n(E) : n(S) - n(E) \), where \( n(E) \) is the number of favorable outcomes and \( n(S) \) is the total number of outcomes. Odds are different from probability.

Question 8. A letter is taken at random from the letters of the word 'ASSISTANT' and another letter is taken at random from the letters of the word 'STATISTICS'. The probability that the selected letters are the same is
(a) \( \frac{7}{45} \)
(b) \( \frac{19}{90} \)
(c) \( \frac{29}{90} \)
(d) \( \frac{19}{90} \)
Answer: (d) \( \frac{19}{90} \)
In simple words: We pick one letter from the word 'ASSISTANT' and another letter from 'STATISTICS'. We want to find the chance that both letters we picked are exactly the same. We need to count the total possible combinations of picking letters, and then count only the combinations where both chosen letters match (e.g., A and A, S and S, T and T, I and I).

🎯 Exam Tip: When dealing with letters in a word, identify the unique letters and their frequencies. The total number of ways to pick letters is the product of the number of letters in each word.

Question 9. A matrix is chosen at random from a set of all matrices of order 2, with elements 0 or 1 only. The probability that the determinant of the matrix chosen is non zero will be
(a) \( \frac{3}{16} \)
(b) \( \frac{3}{8} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{5}{8} \)
Answer: (b) \( \frac{3}{8} \)
In simple words: We are making a 2x2 matrix using only numbers 0 or 1. There are 16 possible matrices we can make. We want to find the chance that the determinant of this matrix is not zero. A non-zero determinant means the matrix is invertible. We list all possible matrices and check their determinants.

🎯 Exam Tip: For a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), the determinant is \( ad - bc \). For a determinant to be non-zero, \( ad \neq bc \).

Question 10. A bag contains 5 white and 3 black balls. Five balls are drawn successively without replacement. The probability that they are alternately of different colours is
(a) \( \frac{3}{14} \)
(b) \( \frac{5}{14} \)
(c) \( \frac{1}{14} \)
(d) \( \frac{9}{14} \)
Answer: (c) \( \frac{1}{14} \)
In simple words: We have a bag with white and black balls. We take out five balls one by one without putting them back. We want to find the chance that the balls we draw keep changing color (e.g., white, then black, then white, etc.). This means there are two possible patterns of drawing colors.

🎯 Exam Tip: When drawing without replacement, the total number of items and the number of specific items decrease with each draw. List all possible alternating patterns and calculate their probabilities separately.

Question 11. If A and B are two events such that \( A \subset B \) and \( P(B) \neq 0 \), then which of the following is correct?
(a) \( P(A/B) = \frac{P(A \cap B)}{P(\overline{B})} \)
(b) \( P(A/B) < P(A) \)
(c) \( P(A/B) \ge P(A) \)
(d) \( P(A/B) > P(A) \)
Answer: (c) \( P(A/B) \ge P(A) \)
In simple words: We have two events, A and B, where A is a part of B (meaning if A happens, B must also happen). We are looking at how the probability of A happening given that B has already happened compares to the probability of A happening on its own. Since A is inside B, knowing B happened makes A more likely, or at least not less likely.

🎯 Exam Tip: If \( A \subset B \), then \( A \cap B = A \). This means \( P(A/B) = \frac{P(A)}{P(B)} \). Since \( P(B) \le 1 \), then \( \frac{1}{P(B)} \ge 1 \), which implies \( \frac{P(A)}{P(B)} \ge P(A) \), or \( P(A/B) \ge P(A) \).

Question 12. A bag contains 6 green, 2 white, and 7 black balls. If two balls are drawn simultaneously, then the probability that both are different colours is
(a) \( \frac{68}{105} \)
(b) \( \frac{64}{105} \)
(c) \( \frac{64}{105} \)
(d) \( \frac{73}{105} \)
Answer: (a) \( \frac{68}{105} \)
In simple words: We have a bag with balls of three different colors. We take out two balls at the same time. We want to find the chance that these two balls are not the same color. We can either calculate the chance of drawing two balls of the same color and subtract it from 1, or sum up the probabilities of drawing (green and white), (green and black), and (white and black).

🎯 Exam Tip: The probability of two balls being of different colors is \( 1 - P(\text{both are same color}) \). The cases for same color are (both green), (both white), or (both black). Calculate these using combinations.

Question 13. If X and Y be two events such that \( P(X/Y) = \frac{1}{2} \), \( P(Y/X) = \frac{1}{3} \) and \( P(X \cap Y) = \frac{1}{6} \), then
(a) \( \frac{2}{5} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{2}{3} \)
Answer: (d) \( \frac{2}{3} \)
In simple words: We are given some conditional probabilities and the probability of both events X and Y happening. We need to find the probability of event X OR event Y happening. We can use the formula for conditional probability to find the individual probabilities of X and Y first. Then use the union formula.

🎯 Exam Tip: Remember the formula for conditional probability: \( P(A/B) = \frac{P(A \cap B)}{P(B)} \). Also, the formula for the union of two events is \( P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) \).

Question 14. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. The probability that the second ball drawn is red will be
(a) \( \frac{5}{12} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{5}{12} \)
(d) \( \frac{5}{12} \)
Answer: (b) \( \frac{1}{2} \)
In simple words: We start with a jar of red and black balls. We pick one ball, see its color, and put it back. Then, we add two more balls of the same color as the one we just picked into the jar. After that, we draw a second ball. We want to find the overall chance that this second ball is red. We consider two scenarios: first ball was red, or first ball was black.

🎯 Exam Tip: This is a classic example of total probability. Break down the problem into mutually exclusive cases (e.g., the first ball drawn was red, or the first ball drawn was black) and sum their probabilities for the desired outcome.

Question 15. A number x is chosen at random from the first 100 natural numbers. Let A be the event of numbers which satisfies \( \frac{(x-10)(x-50)}{x-30} > 0 \) then P(A) is
(1) 0.20
(2) 0.51
(3) 0.71
(4) 0.70
Answer: (3) 0.71
In simple words: We are looking for numbers that make the given fraction positive. This happens when the numerator and denominator are both positive, or both negative. If we check the numbers from 1 to 100, we find that 71 numbers fit this rule. Since there are 100 total numbers, the chance is 71 out of 100.

🎯 Exam Tip: When solving inequalities with fractions, remember to consider the signs of both the numerator and the denominator, and always exclude values that make the denominator zero.

Question 16. If two events A and B are independent such that P(A) = 0.35 and P(A ∪ B) = 0.6, then P(B) is
(1) \( \frac{5}{13} \)
(2) \( \frac{1}{13} \)
(3) \( \frac{4}{13} \)
(4) \( \frac{7}{13} \)
Answer: (1) \( \frac{5}{13} \)
In simple words: Since events A and B are independent, the probability of both happening is just their individual probabilities multiplied together. We use the formula that connects the probabilities of A, B, and A or B happening. By plugging in the given numbers and solving, we can find the probability of event B.

🎯 Exam Tip: Remember that for independent events, the probability of their intersection \( P(A \cap B) \) simplifies to \( P(A) \times P(B) \), which is key to solving such problems.

Question 17. If two events A and B are such that \( P(\overline{A}) = \frac{3}{10} \) and \( P(A \cap B) = \frac{1}{2} \) then \( P(A \cap \overline{B}) \) is
(1) \( \frac{1}{2} \)
(2) \( \frac{1}{3} \)
(3) \( \frac{1}{4} \)
(4) \( \frac{1}{5} \)
Answer: (4) \( \frac{1}{5} \)
In simple words: We know the chance of A not happening and the chance of both A and B happening. We first find the chance of A happening. Then, we can calculate the chance of A happening while B does not, by understanding how these events overlap and subtract. This tells us the probability of A occurring without B.

🎯 Exam Tip: Always use the complement rule \( P(A) = 1 - P(\overline{A}) \) to find the probability of an event if its complement's probability is given.

Question 18. If A and B are two events such that \( P(A) = 0.4 \), \( P(B) = 0.8 \) and \( P(B/A) = 0.6 \), then \( P(\overline{A} \cap B) \) is
(1) 0.96
(2) 0.24
(3) 0.56
(4) 0.66
Answer: (3) 0.56
In simple words: We are given the probability of A, B, and B happening given A. We first use the conditional probability formula to find the chance of both A and B happening. Then, we use this to find the chance of B happening while A does not, by subtracting the overlap from the total probability of B.

🎯 Exam Tip: Remember the formula for conditional probability \( P(B/A) = \frac{P(A \cap B)}{P(A)} \) and use it to find the intersection \( P(A \cap B) \) first.

Question 19. There are three events A, B and C of which one and only one can happen. If the odds are 7 to 4 against A and 5 to 3 against B, then odds against C is
(1) 23 : 65
(2) 65 : 23
(3) 23 : 88
(4) 88 : 23
Answer: (2) 65 : 23
In simple words: "Odds against" an event means the ratio of the probability it won't happen to the probability it will. We use these ratios for A and B to find their exact probabilities. Since only one of A, B, or C can happen, and one of them must, their probabilities add up to 1. We find the probability of C, and then calculate the odds against C.

🎯 Exam Tip: Convert "odds against" into probabilities using the formula: if odds against event E are \( m:n \), then \( P(E) = \frac{n}{m+n} \).

Question 20. If a and b are chosen randomly from the set {1, 2, 3, 4} with replacement, then the probability that the quadratic equation \( x^2 + ax + b = 0 \) has real roots is
(1) \( \frac{3}{16} \)
(2) \( \frac{3}{16} \)
(3) \( \frac{3}{16} \)
(4) \( \frac{3}{16} \)
Answer: (3) \( \frac{3}{16} \)
In simple words: For a quadratic equation to have real roots, a special number called the discriminant must be zero or more. We check all possible pairs of 'a' and 'b' from the given numbers (1, 2, 3, 4). There are 16 such pairs in total. We find which values of 'a' lead to the discriminant being zero or positive. Based on the specified count, there are 3 such cases for 'a' that satisfy this condition, making the probability 3 out of 16.

🎯 Exam Tip: A quadratic equation \( Ax^2 + Bx + C = 0 \) has real roots if its discriminant \( D = B^2 - 4AC \) is greater than or equal to zero.

Question 21. It is given that the events A and B are such that \( P(A) = \frac{1}{4} \), \( P(A/B) = \frac{1}{2} \) and \( P(B/A) = \frac{2}{3} \). Then P(B) is
(1) \( \frac{1}{2} \)
(2) \( \frac{1}{2} \)
(3) \( \frac{1}{2} \)
(4) \( \frac{1}{2} \)
Answer: (2) \( \frac{1}{2} \)
In simple words: We use the rule of conditional probability, which tells us how likely an event is if another event has already happened. We have two conditional probabilities given, along with the probability of event A. By combining these facts through the formulas, we can solve for the probability of event B.

🎯 Exam Tip: Remember the two forms of conditional probability: \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) and \( P(B/A) = \frac{P(A \cap B)}{P(A)} \). These can be used to find unknown probabilities or intersections.

Question 22. In a certain college, 4 % of the boys and 1% of the girls are taller than 1.8 meters. Further 60 % of the students are girls. If a student is selected at random and is taller than 1.8 meters, then the probability that the student is a girl is
(2) \( \frac{3}{11} \)
(3) \( \frac{5}{11} \)
(4) \( \frac{7}{11} \)
Answer: (2) \( \frac{3}{11} \)
In simple words: We want to find the chance that a tall student is a girl. We know the percentage of tall boys and tall girls, and the percentage of girls in the college. We use a formula called Bayes' Theorem, which helps us update probabilities based on new information (like knowing the student is tall). This allows us to figure out the specific probability.

🎯 Exam Tip: This is a classic Bayes' Theorem problem. Clearly define your events (Boy, Girl, Tall) and write down all given probabilities before applying the formula.

Question 23. Ten coins are tossed. The probability of getting at least 8 heads is
(1) \( \frac{7}{64} \)
(2) \( \frac{7}{32} \)
(3) \( \frac{7}{16} \)
(4) \( \frac{7}{128} \)
Answer: (4) \( \frac{7}{128} \)
In simple words: When you toss ten coins, each one can land heads or tails. "At least 8 heads" means you can get exactly 8 heads, 9 heads, or 10 heads. We calculate how many ways each of these can happen and add them up. Then we divide this by the total number of ways the ten coins can land to find the probability.

🎯 Exam Tip: For problems involving "at least X" successes in N trials, calculate the probabilities for X, X+1, ..., N successes and sum them up. Alternatively, calculate \( 1 - P(\text{less than X successes}) \).

Question 24. The probability of two events A and B are 0.3 and 0.6 respectively. The probability that both A and B occur simultaneously is 0.18. The probability that neither A nor B occurs is
(1) 0.1
(2) 0.72
(3) 0.42
(4) 0.28
Answer: (4) 0.28
In simple words: We are given the chances of A happening, B happening, and both A and B happening together. We want to find the chance that neither A nor B happens. First, we figure out the chance of A or B (or both) happening. Then, we subtract this from 1 (the total probability) to get the chance that neither occurs.

🎯 Exam Tip: Remember De Morgan's Law for probabilities: \( P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B) \). This helps find the probability of neither event occurring.

Question 25. If m is a number such that \( m \leq 5 \), then the probability that quadratic equation \( 2x^2 + 2mx + m + 1 = 0 \) has real roots is
(1) \( \frac{1}{5} \)
(2) \( \frac{2}{5} \)
(3) \( \frac{3}{5} \)
(4) \( \frac{4}{5} \)
Answer: (3) \( \frac{3}{5} \)
In simple words: For a quadratic equation to have real answers, a value called the discriminant must be zero or positive. We calculate this discriminant using the 'm' value. We then test the possible integer values for 'm' (from 1 to 5, as it is a number up to 5) to see which ones make the discriminant zero or positive. Out of 5 possible 'm' values, 3 of them work, so the chance is 3 out of 5.

🎯 Exam Tip: When dealing with real roots of a quadratic equation \( ax^2 + bx + c = 0 \), the condition is always \( b^2 - 4ac \geq 0 \). Pay close attention to the range of possible values for any variable like 'm'.