CBSE Class 12 Mathematics Probability MCQs Set 07

Question. The probability that A speaks the truth is \( \frac{4}{5} \) and that of B speaking the truth is \( \frac{3}{4} \). The probability that they contradict each other in stating the same fact is
(a) \( \frac{7}{20} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{3}{20} \)
(d) \( \frac{4}{5} \)
Answer: (a) \( \frac{7}{20} \)

Question. Five fair coins are tossed simultaneously. The probability of the events that atleast one head comes up is
(a) \( \frac{27}{32} \)
(b) \( \frac{5}{32} \)
(c) \( \frac{31}{32} \)
(d) \( \frac{1}{32} \)
Answer: (c) \( \frac{31}{32} \)

Question. A problem in Mathematics is given to three students whose chances of solving it are \( \frac{1}{2}, \frac{1}{3}, \frac{1}{4} \), respectively. If the events of their solving the problem are independent, then the probability that the problem will be solved, is
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{4} \)
Answer: (d) \( \frac{3}{4} \)

Question. A bag contains 3 white, 4 black and 2 red balls. If 2 balls are drawn at random (without replacement), then the probability that both the balls are white, is
(a) \( \frac{1}{18} \)
(b) \( \frac{1}{36} \)
(c) \( \frac{1}{12} \)
(d) \( \frac{1}{24} \)
Answer: (c) \( \frac{1}{12} \)

Question. From the set \( \{1, 2, 3, 4, 5\} \), two numbers a and b \( (a \neq b) \) are chosen at random. The probability that \( \frac{a}{b} \) is an integer, is
(a) \( \frac{1}{3} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{5} \)
Answer: (c) \( \frac{1}{2} \)

Question. An urn contains 6 balls of which two are red and four are black. Two balls are drawn at random. Probability that they are of the different colours is
(a) \( \frac{2}{5} \)
(b) \( \frac{8}{15} \)
(c) \( \frac{7}{15} \)
(d) \( \frac{4}{15} \)
Answer: (b) \( \frac{8}{15} \)

Question. Three dice are thrown simultaneously. The probability of obtaining a total score of 5 is
(a) \( \frac{5}{216} \)
(b) \( \frac{1}{6} \)
(c) \( \frac{1}{36} \)
(d) \( \frac{1}{49} \)
Answer: (c) \( \frac{1}{36} \)

Question. A die is thrown once. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then, \( P(A \cup B) \) is
(a) \( \frac{2}{5} \)
(b) \( \frac{3}{5} \)
(c) 0
(d) 1
Answer: (d) 1

Question. If A and B are events such that \( P\left(\frac{A}{B}\right) = P\left(\frac{B}{A}\right) \neq 0 \), then
(a) \( A \subset B \text{, but } A \neq B \)
(b) \( A = B \)
(c) \( A \cap B = \phi \)
(d) \( P(A) = P(B) \)
Answer: (d) \( P(A) = P(B) \)

Question. If \( P(A \cap B) = \frac{1}{8} \) and \( P(\overline{A}) = \frac{3}{4} \), then \( P\left(\frac{\overline{B}}{\overline{A}}\right) \) is equal to
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{2}{3} \)
Answer: (a) \( \frac{1}{2} \)

Question. For two events A and B, if \( P(A) = 0.4 \), \( P(B) = 0.8 \) and \( P(B / A) = 0.6 \), then \( P(A \cup B) \) is equal to
(a) 0.24
(b) 0.3
(c) 0.48
(d) 0.96
Answer: (d) 0.96

Question. If the sum of numbers obtained on throwing a pair of dice is 9, then the probability that number obtained on one of the dice is 4, is
(a) \( \frac{1}{9} \)
(b) \( \frac{4}{9} \)
(c) \( \frac{1}{18} \)
(d) \( \frac{1}{2} \)
Answer: (d) \( \frac{1}{2} \)

Question. For any two events A and B, if \( P(\overline{A}) = \frac{1}{2} \), \( P(B) = \frac{2}{3} \) and \( P(A \cap B) = \frac{1}{4} \), then \( P\left(\frac{\overline{A}}{\overline{B}}\right) \) is equal to
(a) \( \frac{3}{8} \)
(b) \( \frac{5}{8} \)
(c) \( \frac{1}{8} \)
(d) \( \frac{1}{4} \)
Answer: (b) \( \frac{5}{8} \)

Question. A and B are two events such that \( P(A \cup B) = \frac{3}{4} \), \( P(A \cap B) = \frac{1}{4} \), and \( P(\overline{A}) = \frac{2}{3} \), then \( P(\overline{A} \cap B) \) is equal to
(a) \( \frac{5}{12} \)
(b) \( \frac{3}{8} \)
(c) \( \frac{5}{8} \)
(d) \( \frac{7}{8} \)
Answer: (a) \( \frac{5}{12} \)

Question. A coin is tossed and a card is selected at random from a well shuffled pack of 52 playing cards. The probability of getting head on the coin and a face card from the pack is
(a) \( \frac{3}{12} \)
(b) \( \frac{3}{26} \)
(c) \( \frac{19}{26} \)
(d) \( \frac{3}{13} \)
Answer: (b) \( \frac{3}{26} \)

Question. If A and B are two events such that \( P(A) = 0.2 \), \( P(B) = 0.4 \) and \( P(A \cup B) = 0.5 \), then value of \( P(A / B) \) is
(a) 0.1
(b) 0.25
(c) 0.5
(d) 0.08
Answer: (b) 0.25

Question. If A and B are two events such that \( P(B) = \frac{1}{5} \), \( P\left(\frac{A}{B}\right) = \frac{2}{3} \) and \( P(A \cup B) = \frac{3}{5} \), then \( P(A) \) is
(a) \( \frac{10}{15} \)
(b) \( \frac{2}{15} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{8}{15} \)
Answer: (d) \( \frac{8}{15} \)

Question. If \( P\left(\frac{A}{B}\right) = 0.3 \), \( P(A) = 0.4 \) and \( P(B) = 0.8 \), then \( P\left(\frac{B}{A}\right) \) is equal to
(a) 0.6
(b) 0.3
(c) 0.06
(d) 0.4
Answer: (a) 0.6

Question. The events E and F are independent. If \( P(E) = 0.3 \) and \( P(E \cup F) = 0.5 \), then \( P(E/F) - P(F/E) \) is equal to
(a) \( \frac{1}{7} \)
(b) \( \frac{2}{7} \)
(c) \( \frac{3}{35} \)
(d) \( \frac{1}{70} \)
Answer: (d) \( \frac{1}{70} \)

Question. The probability that A speaks truth is \( \frac{3}{5} \) and for B is \( \frac{3}{4} \), then probability that they contradict each other when asked to speak on a fact.
(a) \( \frac{4}{5} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{9}{20} \)
(d) \( \frac{3}{20} \)
Answer: (c) \( \frac{9}{20} \)

Question. It is given that events A and B are such that \( P(A) = \frac{1}{4} \), \( P\left(\frac{A}{B}\right) = \frac{1}{2} \) and \( P\left(\frac{B}{A}\right) = \frac{2}{3} \), then \( P(B) \) is equal to
(a) \( \frac{2}{3} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{1}{3} \)
Answer: (d) \( \frac{1}{3} \)

Question. A mapping is selected at random from set A={1, 2, ..., 20} into itself. The probability that mapping selected is an injective, is
(a) \( \frac{20}{20^{20}} \)
(b) \( \frac{19!}{20^{19}} \)
(c) \( \frac{19}{20} \)
(d) None of the options
Answer: (b) \( \frac{19!}{20^{19}} \)

Question. If A and B are two events such that \( P(A) = \frac{1}{3} \), \( P(B) = \frac{1}{4} \) and \( P\left(\frac{A}{B}\right) = \frac{1}{5} \), then \( P(\overline{A} \cap \overline{B}) \) is equal to
(a) \( \frac{1}{12} \)
(b) \( \frac{3}{4} \)
(c) \( \frac{7}{15} \)
(d) \( \frac{3}{16} \)
Answer: (c) \( \frac{7}{15} \)

Question. The chances that doctor A will diagnose a disease X correctly is 60%. The chances that a patient will die by his treatment after correct diagnose is 40% and the chance of death by wrong diagnose is 70%. A patient of doctor A, who had disease X, died. What is the chance that his disease was diagnosed correctly?
(a) \( \frac{7}{13} \)
(b) \( \frac{3}{10} \)
(c) \( \frac{6}{13} \)
(d) \( \frac{7}{10} \)
Answer: (c) \( \frac{6}{13} \)

Assertion-Reason Based Questions

Question. Two coins are tossed once.
Assertion (A) If E: tail appears on one coin and F: one coin shows head, then \( P(E/F) \) is 1.
Reason (R) If E: no tail appears and F: no head appears, then \( P(E/F) \) is 0.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (b) Both A and R are correct; R is not the correct explanation of A.

Question. Assertion (A) Two coins are tossed simultaneously. The probability getting two heads, if it is known that atleast one head comes up, is \( \frac{1}{3} \).
Reason (R) Let E and F be two events with a random experiment, \( P\left(\frac{F}{E}\right) = \frac{P(E \cap F)}{P(E)} \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A) If A and B are two events such that \( 0 < P(A), P(B) < 1 \), then \( P(A / B) + P(\overline{A} / B) = 3/2 \).
Reason (R) If A and B are two events such that \( 0 < P(A), P(B) < 1 \), then \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) and \( P(B) = P(A \cap B) + P(\overline{A} \cap B) \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (d) R is correct; A is incorrect.

Question. Assertion (A) Three cards are drawn from a well shuffled deck of 52 cards without replacement, then the probability that first two cards are black and third card is red is \( \frac{13}{102} \).
Reason (R) For three events, \( P(A \cap B \cap C) = P(A) \cdot P\left(\frac{B}{A}\right) \cdot P\left(\frac{C}{A \cap B}\right) \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A) Consider the experiment of drawing a card from a deck of 52 playing cards, in which the elementary events are assumed to be equally likely. If E and F denote the events the card drawn is a spade and the card drawn is an ace respectively, then \( P(E/F) = \frac{1}{4} \) and \( P(F/E) = \frac{1}{13} \).
Reason (R) E and F are two events such that the probability of occurrence of one of them is not affected by occurrence of the other. Such events are called independent events.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (b) Both A and R are correct; R is not the correct explanation of A.

Question. Assertion (A) If A and B are independent events such that \( P(A) = \frac{3}{5} \) and \( P(B) = \frac{1}{5} \), then \( P(A \cap B) \) is \( \frac{3}{25} \).
Reason (R) Two cards are drawn at random and without replacement from a pack of 52 playing cards. Then, the probability that both the cards are red, is \( \frac{25}{102} \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (b) Both A and R are correct; R is not the correct explanation of A.

Question. Assertion (A) In rolling a dice, event A = {1, 3, 5} and event B = {2, 4} are mutually exclusive events.
Reason (R) In a sample space, two events are mutually exclusive if they do not occur at the same time.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A) Let A and B be two independent events and \( P(A) = 0.3, P(A \cup \overline{B}) = 0.8 \), then \( P(B) = \frac{2}{7} \).
Reason (R) \( P(\overline{E}) = 1 - P(E) \), where E be any event.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (b) Both A and R are correct; R is not the correct explanation of A.

Question. Assertion (A) Three coins are tossed, consider the event E 'three heads or three tails' and F 'atmost two heads'. Events E and F are independent.
Reason (R) Two events are independents if \( P(A \cap B) = P(A) \cdot P(B) \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (d) R is correct; A is incorrect.

Case Study Based Questions

Case Study 1:
There are different types of Yoga which involve the usage of different poses of Yoga Asanas, Meditation and Pranayam. The Venn diagram below represents the probabilities of three different types of Yoga, A, B and C performed by the people of a society. Further, it is given that probability of a member performing type C Yoga is 0.44.

• Circle A (left) and Circle B (middle) overlap at 0.09.
• Circle B (middle) and Circle C (right) overlap at x.
• Circle B has an exclusive region labeled y.
• Circle C has an exclusive region labeled 0.21.
• Circle A has an exclusive region labeled 0.32.
• The region outside all three circles is 0.11.

Question. Find the value of x.
Answer: From the Venn diagram, we have: \[ P(C) = x + 0.21 \] Given that the probability of a member performing type C Yoga is 0.44, we can set up the equation: \[ x + 0.21 = 0.44 \] \[ x = 0.44 - 0.21 = 0.23 \] Therefore, the value of \( x \) is 0.23.

Question. Find the value of y.
Answer: The total probability of all mutually exclusive regions in the sample space \( S \) must sum to 1. From the Venn diagram, these regions are: \[ 0.32 + 0.09 + y + x + 0.21 + 0.11 = 1 \] Substituting \( x = 0.23 \) into the equation: \[ 0.32 + 0.09 + y + 0.23 + 0.21 + 0.11 = 1 \] \[ 0.96 + y = 1 \] \[ y = 1 - 0.96 = 0.04 \] Therefore, the value of \( y \) is 0.04.

Question. Find \( P\left(\frac{C}{B}\right) \).
Answer: By definition of conditional probability: \[ P\left(\frac{C}{B}\right) = \frac{P(C \cap B)}{P(B)} \] From the Venn diagram: \[ P(C \cap B) = x = 0.23 \] \[ P(B) = 0.09 + y + x = 0.09 + 0.04 + 0.23 = 0.36 \] Therefore: \[ P\left(\frac{C}{B}\right) = \frac{0.23}{0.36} = \frac{23}{36} \]

Question. Find the probability that a randomly selected person of the society does yoga of type A or B but not C.
Answer: The event "does yoga of type A or B but not C" corresponds to the region \( (A \cup B) \setminus C \). From the Venn diagram, this region consists of: - The exclusive region of A: \( 0.32 \) - The intersection of A and B: \( 0.09 \) - The exclusive region of B: \( y = 0.04 \) Therefore, the probability is: \[ 0.32 + 0.09 + 0.04 = 0.45 \]

Case Study 2:
There are two anti-aircraft guns, named as A and B. The probabilities that the shell fired from them hits an airplane are 0.3 and 0.2, respectively. Both of them fired one shell at an airplane at the same time.

Question. What is the probability that the shell fired from exactly one of them hit the plane?
Answer: Let \( P(A) = 0.3 \) be the probability that gun A hits the plane, and \( P(B) = 0.2 \) be the probability that gun B hits the plane. The probability that gun A does not hit the plane is \( P(\overline{A}) = 1 - 0.3 = 0.7 \), and that gun B does not hit the plane is \( P(\overline{B}) = 1 - 0.2 = 0.8 \). The probability that exactly one of them hits the plane is: \[ P(\text{exactly one hits}) = P(A) \cdot P(\overline{B}) + P(\overline{A}) \cdot P(B) \] \[ = 0.3 \times 0.8 + 0.7 \times 0.2 \] \[ = 0.24 + 0.14 = 0.38 \]

Question. If it is known that the shell fired from exactly one of them hit the plane, then what is the probability that it was fired from B?
Answer: The probability that only B hit the plane is: \[ P(\overline{A} \cap B) = P(\overline{A}) \cdot P(B) = 0.7 \times 0.2 = 0.14 \] The probability that exactly one of them hit the plane is \( 0.38 \). Using Bayes' theorem, the required conditional probability is: \[ P(\text{fired from B} \mid \text{exactly one hit}) = \frac{P(\overline{A} \cap B)}{P(\text{exactly one hit})} = \frac{0.14}{0.38} = \frac{7}{19} \]

Case Study 3:
In a clinic, out of 40 patients who are waiting to see the doctor:

• 24 have cold
• 18 have fever
• 9 have migraine
• 4 have both cold and fever
• 6 have both fever and migraine
• 3 have both migraine and cold
• 2 have all the three

(Note: Assume that each of the 40 patients present is suffering from one or more of the 3 ailments - cold, fever or migraine only.)

Question. If a patient is selected at random, find the probability that he/she does not have migraine given that he/she has fever. Show your work.
Answer: Let \( F \) be the event that the patient has fever and \( M \) be the event that the patient has migraine. From the given data: - Total patients = 40 - Number of patients with fever, \( n(F) = 18 \implies P(F) = \frac{18}{40} \) - Number of patients with both fever and migraine, \( n(M \cap F) = 6 \implies P(M \cap F) = \frac{6}{40} \) The probability that the patient does not have migraine given that he/she has fever is: \[ P(\overline{M} \mid F) = 1 - P(M \mid F) = 1 - \frac{P(M \cap F)}{P(F)} \] \[ = 1 - \frac{6/40}{18/40} = 1 - \frac{6}{18} = 1 - \frac{1}{3} = \frac{2}{3} \]

Question. If a patient is selected at random, what is the probability that he/she has cold or fever or both, given that he/she has migraine? Show your work.
Answer: Let \( C \), \( F \), and \( M \) represent the events that a patient has cold, fever, and migraine respectively. From the given data: - Total patients = 40 - Number of patients with migraine, \( n(M) = 9 \) - Number of patients with both migraine and cold, \( n(C \cap M) = 3 \) - Number of patients with both fever and migraine, \( n(F \cap M) = 6 \) - Number of patients with all three ailments, \( n(C \cap F \cap M) = 2 \) The probability that a patient has cold or fever or both, given that he/she has migraine is \( P(C \cup F \mid M) \), which can be calculated as: \[ P(C \cup F \mid M) = P(C \mid M) + P(F \mid M) - P(C \cap F \mid M) \] Where: \[ P(C \mid M) = \frac{P(C \cap M)}{P(M)} = \frac{3}{9} = \frac{1}{3} \] \[ P(F \mid M) = \frac{P(F \cap M)}{P(M)} = \frac{6}{9} = \frac{2}{3} \] \[ P(C \cap F \mid M) = \frac{P(C \cap F \cap M)}{P(M)} = \frac{2}{9} \] Substituting these values: \[ P(C \cup F \mid M) = \frac{1}{3} + \frac{2}{3} - \frac{2}{9} = 1 - \frac{2}{9} = \frac{7}{9} \]

Case Study 4:
A departmental store sends bills to charge its customers once a month. Past experience shows that 70% of its customers pay their first month bill in time. The store also found that the customer who pays the bill in time has the probability of 0.8 of paying in time next month and the customer who does not pay in time has the probability of 0.4 of paying in time the next month.

Question. Let \( E_1 \) and \( E_2 \) respectively denote the event of customer paying or not paying the first month bill in time. Find \( P(E_1) \) and \( P(E_2) \).
Answer: Since 70% of the customers pay their first month bill in time, we have: \[ P(E_1) = 70\% = 0.7 \] The probability of customers not paying the first month bill in time is: \[ P(E_2) = 1 - P(E_1) = 1 - 0.7 = 0.3 \]

Question. Let A denotes the event of customer paying second month’s bill in time, then find \( P\left(\frac{A}{E_1}\right) \) and \( P\left(\frac{A}{E_2}\right) \).
Answer: From the given information: - The probability that a customer who paid the first month bill in time also pays the second month bill in time is: \[ P\left(\frac{A}{E_1}\right) = 0.8 \] - The probability that a customer who did not pay the first month bill in time pays the second month bill in time is: \[ P\left(\frac{A}{E_2}\right) = 0.4 \Workspace \]

Question. Find the probability of customer paying second month’s bill in time.
Answer: Using the law of total probability: \[ P(A) = P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right) \] \[ = 0.7 \times 0.8 + 0.3 \times 0.4 \] \[ = 0.56 + 0.12 = 0.68 \]

Question. Find the probability of customer paying first month's bill in time, if it is found that customer has paid the second month's bill in time.
Answer: Using Bayes' theorem: \[ P(E_1 \mid A) = \frac{P(E_1) \cdot P\left(\frac{A}{E_1}\right)}{P(A)} \] \[ = \frac{0.7 \times 0.8}{0.68} \] \[ = \frac{0.56}{0.68} = \frac{14}{17} \approx 0.824 \]

Case Study 5:
A building contractor undertakes a job to construct 4 flats on a plot along with parking area. Due to strike, the probability of many construction workers not being present for the job is 0.65. The probability that many are not present and still the work gets completed on time is 0.35. The probability that work will be completed on time when all workers are present is 0.80.
Let \( E_1 \) : represents the event when many workers were not present for the job; \( E_2 \) : represents the event when all workers were present; and \( E \) : represents completing the construction work on time.

Question. What is the probability that all the workers are present for the job?
Answer: Since \( E_1 \) is the event that many workers are not present and \( E_2 \) is the event that all workers are present: \[ P(E_2) = 1 - P(E_1) = 1 - 0.65 = 0.35 \]

Question. What is the probability that construction will be completed on time?
Answer: We are given: - \( P(E_1) = 0.65 \) - \( P(E \mid E_1) = 0.35 \) - \( P(E \mid E_2) = 0.80 \) - From the previous question, \( P(E_2) = 0.35 \). Using the law of total probability: \[ P(E) = P(E_1) \cdot P(E \mid E_1) + P(E_2) \cdot P(E \mid E_2) \] \[ = 0.65 \times 0.35 + 0.35 \times 0.80 \] \[ = 0.2275 + 0.28 = 0.5075 \approx 0.51 \]

Question. What is the probability that many workers are not present given that the construction work is completed on time?
Answer: Using Bayes' theorem: \[ P(E_1 \mid E) = \frac{P(E_1) \cdot P(E \mid E_1)}{P(E)} \] \[ = \frac{0.65 \times 0.35}{0.5075} = \frac{0.2275}{0.5075} \approx 0.45 \]

Question. What is the probability that all workers were present given that the construction job was completed on time?
Answer: Using Bayes' theorem: \[ P(E_2 \mid E) = \frac{P(E_2) \cdot P(E \mid E_2)}{P(E)} \] \[ = \frac{0.35 \times 0.80}{0.5075} = \frac{0.28}{0.5075} \approx 0.55 \]

Case Study 6:
Recent studies suggest that roughly 12% of the world population is left handed. Depending upon the parents, the chances of having a left handed child are as follows:
A: When both father and mother are left handed: Chances of left handed child is 24%.
B: When father is right handed and mother is left handed: Chances of left handed child is 22%.
C: When father is left handed and mother is right handed: Chances of left handed child is 17%.
D: When both father and mother are right handed: Chances of left handed child is 9%.
Assuming that \( P(A) = P(B) = P(C) = P(D) = \frac{1}{4} \) and \( L \) denotes the event that child is left handed.

Question. Find \( P(L/C) \).
Answer: From the given data, when the father is left handed and the mother is right handed (event C), the chance of having a left handed child is 17%. Therefore: \[ P(L \mid C) = 17\% = \frac{17}{100} = 0.17 \]

Question. Find \( P(\overline{L}/A) \).
Answer: First, find the probability of a left-handed child when both parents are left-handed (event A): \[ P(L \mid A) = 24\% = \frac{24}{100} = 0.24 \] The probability that the child is not left-handed given event A is: \[ P(\overline{L} \mid A) = 1 - P(L \mid A) = 1 - 0.24 = 0.76 = \frac{19}{25} \]

Question. Find \( P(A/L) \).
Answer: Using Bayes' theorem: \[ P(A \mid L) = \frac{P(A) \cdot P(L \mid A)}{P(A) \cdot P(L \mid A) + P(B) \cdot P(L \mid B) + P(C) \cdot P(L \mid C) + P(D) \cdot P(L \mid D)} \] Given \( P(A) = P(B) = P(C) = P(D) = \frac{1}{4} \), the prior probabilities cancel out: \[ P(A \mid L) = \frac{P(L \mid A)}{P(L \mid A) + P(L \mid B) + P(L \mid C) + P(L \mid D)} \] Substituting the given values: \[ P(A \mid L) = \frac{0.24}{0.24 + 0.22 + 0.17 + 0.09} = \frac{0.24}{0.72} = \frac{1}{3} \]

Question. Find the probability that a randomly selected child is left handed given that exactly one of the parents is left handed.
Answer: Exactly one of the parents being left-handed corresponds to the union of events \( B \) and \( C \) (i.e., \( B \cup C \)). The required conditional probability is \( P(L \mid B \cup C) \). By definition: \[ P(L \mid B \cup C) = \frac{P(L \cap (B \cup C))}{P(B \cup C)} = \frac{P(L \cap B) + P(L \cap C)}{P(B) + P(C)} \] \[ = \frac{P(B) \cdot P(L \mid B) + P(C) \cdot P(L \mid C)}{P(B) + P(C)} \] Substituting \( P(B) = P(C) = \frac{1}{4} \), we get: \[ P(L \mid B \cup C) = \frac{\frac{1}{4}(0.22) + \frac{1}{4}(0.17)}{\frac{1}{4} + \frac{1}{4}} = \frac{0.22 + 0.17}{2} = \frac{0.39}{2} = 0.195 \text{ (or } \frac{39}{200} \text{)} \]