RS Aggarwal Solutions for Class 12 Chapter 29 Probability

Question 1. Let A and B be the events such that P(A) = 7/13, P(B) = 9/13 and P(A ∩ B) = 4/13. Find (i) P(A / B) (ii) P(B / A) (iii) P(A ∪ B) (iv) P(B̄ / Ā)
Answer: Given - A and B are events with P(A) = 7/13, P(B) = 9/13 and P(A ∩ B) = 4/13.

To find - (i) P(A/B) (ii) P(B/A) (iii) P(A ∪ B) (iv) P(B̄/Ā)

Formula to be used - By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

(i) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{4/13}{9/13} = \frac{4}{9} \)

(ii) \( P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{4/13}{7/13} = \frac{4}{7} \)

(iii) \( P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{7}{13} + \frac{9}{13} - \frac{4}{13} = \frac{12}{13} \)

(iv) \( P(B̄/Ā) = \frac{P(B̄ \cap Ā)}{P(Ā)} \)

Now, by De-Morgan's Law, \( (A \cup B)^C = A^C \cap B^C \)

\( \therefore P(Ā \cap B̄) = P(\overline{A \cup B}) \)

\( \therefore \frac{P(Ā \cap B̄)}{P(Ā)} = \frac{P(\overline{A \cup B})}{P(Ā)} = \frac{1 - P(A \cup B)}{1 - P(A)} = \frac{1 - 12/13}{1 - 7/13} = \frac{1/13}{6/13} = \frac{1}{6} \)

Exam Tip: Use the conditional probability formula carefully and remember De-Morgan's Law when finding probabilities of complements.

Question 2. Let A and B be the events such that P(A) = 5/11, P(B) = 6/11 and P(A ∪ B) = 7/11. Find (i) P(A ∩ B) (ii) P(A / B) (iii) P(B / A) (iv) P(Ā / B̄)
Answer: Given - A and B are events with P(A) = 5/11, P(B) = 6/11 and P(A ∪ B) = 7/11.

To find - (i) P(A ∩ B) (ii) P(A/B) (iii) P(B/A) (iv) P(Ā/B̄)

Formula to be used - By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

(i) \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)

\( \Rightarrow P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{5}{11} + \frac{6}{11} - \frac{7}{11} = \frac{4}{11} \)

(ii) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{4/11}{6/11} = \frac{4}{6} = \frac{2}{3} \)

(iii) \( P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{4/11}{5/11} = \frac{4}{5} \)

(iv) \( P(Ā/B̄) = \frac{P(Ā \cap B̄)}{P(B̄)} \)

Now, by De-Morgan's Law, \( (A \cup B)^C = A^C \cap B^C \)

\( \therefore P(Ā \cap B̄) = P(\overline{A \cup B}) \)

\( \therefore \frac{P(Ā \cap B̄)}{P(B̄)} = \frac{P(\overline{A \cup B})}{P(B̄)} = \frac{1 - P(A \cup B)}{1 - P(B)} = \frac{1 - 7/11}{1 - 6/11} = \frac{4/11}{5/11} = \frac{4}{5} \)

Exam Tip: Always check that P(A ∪ B) matches the formula before solving for P(A ∩ B), as this is the foundation for all conditional probabilities.

Question 3. Let A and B be the events such that P(A) = 3/10, P(B) = 1/2 and P(B / A) = 2/5. Find (i) P(A ∩ B) (ii) P(A ∪ B) (iii) P(A / B)
Answer: Given - A and B are events with P(A) = 3/10, P(B) = 1/2 and P(B/A) = 2/5.

To find - (i) P(A ∩ B) (ii) P(A ∪ B) (iii) P(A/B)

Formula to be used - By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

(i) \( P(B/A) = \frac{P(A \cap B)}{P(A)} \)

\( \Rightarrow P(A \cap B) = P(A) \times P(B/A) = \frac{3}{10} \times \frac{2}{5} = \frac{6}{50} = \frac{3}{25} \)

(ii) \( P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{3}{10} + \frac{1}{2} - \frac{3}{25} = \frac{15 + 25 - 6}{50} = \frac{34}{50} = \frac{17}{25} \)

(iii) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{3/25}{1/2} = \frac{3}{25} \times \frac{2}{1} = \frac{6}{25} \)

Exam Tip: When given P(B/A), multiply it by P(A) to find the intersection - this is the reverse application of the conditional probability formula.

Question 4. Let A and B be the events such that 2P(A) = P(B) = 5/13 and P(A / B) = 2/5. Find (i) P(A ∩ B) (ii) P(A ∪ B)
Answer: Given - A and B are events with 2P(A) = P(B) = 5/13 and P(A/B) = 2/5.

To find - (i) P(A ∩ B) (ii) P(A ∪ B)

Formula to be used - By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

From the given, P(B) = 5/13, so P(A) = 5/26.

(i) \( P(A/B) = \frac{P(A \cap B)}{P(B)} \)

\( \Rightarrow P(A \cap B) = P(B) \times P(A/B) = \frac{5}{13} \times \frac{2}{5} = \frac{2}{13} \)

(ii) \( P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13} = \frac{5}{26} + \frac{10 - 4}{26} = \frac{5 + 6}{26} = \frac{11}{26} \)

Exam Tip: When probabilities are given in ratios like 2P(A) = P(B), extract the individual values first before applying conditional probability formulas.

Question 5. A die is rolled. If the outcome is an even number, what is the probability that it is a number greater than 2?
Answer: A die has 6 faces with sample space S = {1, 2, 3, 4, 5, 6}. The total number of outcomes is 6.

Let P(A) denote the probability of getting an even number. The sample space of A = {2, 4, 6}, so P(A) = 3/6 = 1/2.

Let P(B) denote the probability of getting a number greater than 2. The sample space of B = {3, 4, 5, 6}.

By conditional probability, \( P(B/A) = \frac{P(A \cap B)}{P(A)} \) where P(B/A) is the probability of obtaining an outcome greater than 2 given that the outcome is even.

The event (A ∩ B) = {4, 6}, so \( P(A \cap B) = \frac{2}{6} = \frac{1}{3} \)

Therefore, \( P(B/A) = \frac{1/3}{1/2} = \frac{2}{3} \)

Exam Tip: Identify the given condition (even numbers here) first, then find all outcomes satisfying both the condition and the required event.

Question 6. A coin is tossed twice. If the outcome is at most one tail, what is the probability that both head and tail have appeared?
Answer: A coin has 2 sides with sample space S = {H, T}. The total number of outcomes is 2. A coin is tossed twice.

Let P(A) denote the probability of getting at most 1 tail. The sample space of A = {(H, H), (H, T), (T, H)}.

Let P(B) denote the probability of getting a head. The sample space of B = {H}.

\( \therefore P(B) = \frac{1}{2} \)

The probability of getting at most one tail and a head: (A ∩ B) = {(H, H)}

\( \therefore P(A \cap B) = \frac{1}{3} \)

By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

The probability that both head and tail have appeared: \( P(A/B) = \frac{1/3}{1/2} = \frac{2}{3} \)

Exam Tip: When working with coin tosses, list all possible ordered outcomes carefully - the order matters when distinguishing between "first" and "second" toss.

Question 7. Three coins are tossed simultaneously. Find the probability that all coins show heads if at least one of the coins shows a head.
Answer: When three coins are tossed simultaneously, the total number of outcomes is \( 2^3 = 8 \), and the sample space is S = {(H,H,H), (H,H,T), (H,T,T), (H,T,H), (T,H,T), (T,T,H), (T,H,H), (T,T,T)}.

Let P(A) denote the probability of getting 3 heads. The sample space of A = {(H, H, H)}, so \( P(A) = \frac{1}{8} \).

Let P(B) denote the probability of getting at least one head. The probability of one head = 1 - probability of no heads = 1 - 1/8 = 7/8, so \( P(B) = \frac{7}{8} \).

The probability that the throw yields either all heads or at least one head: \( P(A \cup B) = \frac{7}{8} \)

Now, \( P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{1}{8} + \frac{7}{8} - \frac{7}{8} = \frac{1}{8} \)

By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

The probability that all coins show heads if at least one of the coins showed a head: \( P(A/B) = \frac{1/8}{7/8} = \frac{1}{7} \)

Exam Tip: When finding the complement of an event ("at least one head"), calculate it as 1 minus the probability of the opposite ("no heads").

Question 8. Two unbiased dice are thrown. Find the probability that the sum of the numbers appearing is 8 or greater, if 4 appears on the first die.
Answer: Two dice, each with 6 faces, when tossed simultaneously yield a total of 6² = 36 outcomes.

Let P(A) denote the probability of getting a sum greater than 8. Let P(B) denote the probability of getting 4 on the first die. The sample space of B = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)}.

Let P(A ∩ B) denote the probability of getting 4 on the first die and a sum greater than or equal to 8. The sample space of (A ∩ B) = {(4,4), (4,5), (4,6)}.

\( \therefore P(A \cap B) = \frac{3}{36} = \frac{1}{12} \)

By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

The probability that the sum of the numbers is greater than or equal to 8 given that 4 was thrown first: \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{1/12}{1/6} = \frac{1}{2} \)

Exam Tip: For two-dice problems, count carefully - given that one die shows a fixed value, determine only the favorable outcomes for the other die.

Question 9. A die is thrown twice and the sum of the numbers appearing is observed to be 8. What is the conditional probability that the number 5 has appeared at least once?
Answer: A die thrown twice will have a total of 6² = 36 outcomes.

Let P(A) denote the probability of getting the number 5 at least once. Let P(B) denote the probability of getting a sum equal to 8. The sample space of B = {(2,6), (3,5), (4,4), (5,3), (6,2)}.

\( \therefore P(B) = \frac{5}{36} \)

Let P(A ∩ B) denote the probability of getting the number 5 at least once and a sum equal to 8. The sample space of (A ∩ B) = {(3,5), (5,3)}.

\( \therefore P(A \cap B) = \frac{2}{36} = \frac{1}{18} \)

By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

The probability that the number 5 has appeared at least once given that the sum is 8: \( P(A/B) = \frac{1/18}{5/36} = \frac{2}{5} \)

Exam Tip: List all ordered pairs satisfying the given condition (sum = 8) first, then identify which of these also satisfy the required event (contain 5).

Question 10. Two dice were thrown and it is known that the numbers which come up were different. Find the probability that the sum of the two numbers was 5.
Answer: Two dice, each with 6 faces, when tossed simultaneously yield a total of 6² = 36 outcomes.

Let P(A) denote the probability of getting a sum equal to 5. Let P(B) denote the probability of getting 2 different numbers.

Probability of getting 2 different numbers = 1 - probability of getting same numbers = 1 - 1/6 = 5/6.

Let P(A ∩ B) denote the probability of getting a sum equal to 5 and two different numbers at the same time. The sample space of (A ∩ B) = {(1,4), (2,3), (3,2), (4,1)}.

\( \therefore P(A \cap B) = \frac{4}{36} = \frac{1}{9} \)

By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

The probability that the sum equals 5 given that two different numbers were thrown: \( P(A/B) = \frac{1/9}{5/6} = \frac{2}{15} \)

Exam Tip: Always separate the given condition from the required event - the condition defines which outcomes are possible, and then find how many of those satisfy the event.

Question 11. A coin is tossed and then a die is thrown. Find the probability of obtaining a 6, given that a head came up.
Answer: A coin is tossed and a die is thrown. A coin with two sides has a total of 2 outcomes: {H, T}. A die has 6 faces with a total of 6 outcomes: {1, 2, 3, 4, 5, 6}.

Let P(A) denote the probability of getting the number 6. \( \therefore P(A) = \frac{1}{6} \)

Let P(B) denote the probability of getting a head. The sample space of B = {H}. \( \therefore P(B) = \frac{1}{2} \)

Let P(A ∩ B) denote the probability of getting the number 6 and a head. \( \therefore P(A \cap B) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12} \)

By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

The probability that 6 came up given that a head came up: \( P(A/B) = \frac{1/12}{1/2} = \frac{1}{6} \)

Exam Tip: For independent events like coin toss and die roll, the intersection probability is the product of individual probabilities.

Question 12. A couple has 2 children. Find the probability that both are boys if it is known that (i) one of the children is a boy, and (ii) the elder child is a boy.
Answer: A couple has two children. The sample space S = {(B,B), (B,G), (G,B), (G,G)}.

Let P(A) denote the probability of both being boys.

(i) Let P(B) denote the probability of one of them being a boy. The sample space of B = {(B,B), (B,G), (G,B)}.

\( \therefore P(B) = \frac{3}{4} \)

Let P(A ∩ B) denote the probability of one of them being a boy and both being boys. \( \therefore (A \cap B) = \{(B,B)\} \)

\( \therefore P(A \cap B) = \frac{1}{4} \)

By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

The probability that both are boys given that one of them is a boy: \( P(A/B) = \frac{1/4}{3/4} = \frac{1}{3} \)

(ii) Let P(B) denote the probability of the elder being a boy. The sample space of B = {(B,B), (B,G)}.

\( \therefore P(B) = \frac{2}{4} = \frac{1}{2} \)

Let P(A ∩ B) denote the probability of the elder being a boy and both being boys. \( \therefore (A \cap B) = \{(B,B)\} \)

\( \therefore P(A \cap B) = \frac{1}{4} \)

By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

The probability that both are boys given that the elder is a boy: \( P(A/B) = \frac{1/4}{1/2} = \frac{1}{2} \)

Exam Tip: Notice how the condition changes the answer - "one is a boy" gives 1/3, while "the elder is a boy" gives 1/2. The specificity of the condition matters.

Question 13. In a class, 40% students study mathematics; 25% study biology and 15% study both mathematics and biology. One student is selected at random. Find the probability that (i) he studies mathematics if it is known that he studies biology (ii) he studies biology if it is known that he studies mathematics.
Answer: Let P(A) denote the probability of students studying mathematics. \( \therefore P(A) = 0.40 \)

Let P(B) denote the probability of students studying biology. \( \therefore P(B) = 0.25 \)

Let P(A ∩ B) denote the probability of students studying both mathematics and biology. \( \therefore P(A \cap B) = 0.15 \)

One student is selected at random.

By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

(i) The probability that he studies mathematics given that he studies biology: \( P(A/B) = \frac{0.15}{0.25} = \frac{3}{5} \)

(ii) The probability that he studies biology given that he studies mathematics: \( P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{0.15}{0.40} = \frac{3}{8} \)

Exam Tip: When working with percentages in probability, convert them to decimals or fractions first to avoid calculation errors.

Question 14. The probability that a student selected at random from a class will pass in Hindi is 4/5 and the probability that he passes in Hindi and English is 1/2. What is the probability that he will pass in English if it is known that he has passed in Hindi?
Answer: One student is selected at random. Let P(A) denote the probability of students passing in English. Let P(B) denote the probability of students passing in Hindi. \( \therefore P(B) = \frac{4}{5} \)

Let P(A ∩ B) denote the probability of students passing in both English and Hindi. \( \therefore P(A \cap B) = \frac{1}{2} \)

By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

The probability that he will pass in English given that he passes in Hindi: \( P(A/B) = \frac{1/2}{4/5} = \frac{5}{8} \)

Exam Tip: When dividing fractions, multiply by the reciprocal - this is a common step in conditional probability problems.

Question 15. The probability that a certain person will buy a shirt is 0.2, the probability that he will buy a coat is 0.3 and the probability that he will buy a shirt given that he buys a coat is 0.4. Find the probability that he will buy both a shirt and a coat.
Answer: Let P(A) denote the probability of a certain person buying a shirt. \( \therefore P(A) = 0.2 \)

Let P(B) denote the probability of him buying a coat. \( \therefore P(B) = 0.3 \)

Let P(A ∩ B) denote the probability that he buys both a shirt and a coat.

By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

The probability that he will buy a shirt given that he buys a coat: \( P(A/B) = \frac{P(A \cap B)}{P(B)} = 0.4 \)

\( \Rightarrow P(A \cap B) = P(B) \times P(A/B) = 0.3 \times 0.4 = 0.12 \)

Exam Tip: When finding the intersection given the conditional probability, rearrange the formula to P(A ∩ B) = P(B) × P(A/B).

Question 16. In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. (i) Find the probability that he reads neither Hindi nor English news paper. (ii) If he reads Hindi newspaper, what is the probability that he reads English newspaper? (iii) If he reads English newspaper, what is the probability that he reads Hindi newspaper?
Answer: Let P(A) denote the probability of students reading Hindi newspaper. \( \therefore P(A) = 0.60 \)

Let P(B) denote the probability of them reading English newspaper. \( \therefore P(B) = 0.40 \)

Let P(A ∩ B) denote the probability of them reading both. \( \therefore P(A \cap B) = 0.20 \)

Let P(A ∪ B) denote the probability of them reading either one of them. \( \therefore P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.60 + 0.40 - 0.20 = 0.80 \)

(i) The probability that none of them reads either of them = 1 - 0.8 = 0.2 = 1/5

By conditional probability, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) where P(A/B) is the probability of occurrence of event A given that B has already occurred.

(ii) The probability that he reads the English one given that he reads the Hindi one: \( P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{0.20}{0.60} = \frac{1}{3} \)

(iii) The probability that he reads the Hindi one given that he reads the English one: \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0.20}{0.40} = \frac{1}{2} \)

Exam Tip: Part (i) uses the complement rule, while parts (ii) and (iii) use conditional probability - notice which event is the given condition.

Question 17. Two integers are selected at random from integers 1 through 11. If the sum is even, find the probability that both the numbers selected are odd.
Answer: Two integers are selected at random. The first choice has 11 options from the 11 integers, and the second choice has 10 options from the remaining 10 integers.

Let P(A) denote the probability of choosing both numbers odd. Let P(B) denote the probability of choosing the numbers to yield an even sum.

Sample space of B = {(1,3), (1,5), (1,7), (1,9), (1,11), (3,5), (3,7), (3,9), (3,11), (5,7), (5,9), (5,11), (7,9), (7,11), (9,11), (2,4), (2,6), (2,8), (2,10), (4,6), (4,8), (4,10), (6,8), (6,10), (8,10)}

\( \therefore P(B) = \frac{25}{110} \)

Let P(A ∩ B) denote the probability of getting both odd numbers giving an even sum. \( \therefore (A \cap B) = \{(1,3), (1,5), (1,7), (1,9), (1,11), (3,5), (3,7), (3,9), (3,11), (5,7), (5,9), (5,11), (7,9), (7,11), (9,11)\} \)

\( \therefore P(A \cap B) = \frac{15}{110} \)

The probability of getting both numbers odd given that the sum is even: \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{15/110}{25/110} = \frac{15}{25} = \frac{3}{5} \)

Exam Tip: For counting ordered pairs, remember that the first selection affects the number of choices remaining for the second selection.

Question 1. A bag contains 17 tickets, numbered from 1 to 17. A ticket is drawn, and then another ticket is drawn without replacing the first one. Find the probability that both the tickets may show even numbers.
Answer: Given - A bag contains 17 tickets numbered 1 to 17, and each trial is independent of the other. Hence the sample space is given by S = {1, 2, 3, ..., 17}.

To find - the probability that both tickets drawn show even numbers.

Let success denote that the ticket drawn is even, i.e., {2, 4, 6, 8, 10, 12, 14, 16}. There are 8 even-numbered tickets among the 17 total.

Now, the probability of success in the first trial is \( P(\text{first even}) = \frac{8}{17} \)

After removing one even-numbered ticket, 7 even-numbered tickets remain out of 16 total tickets. The probability of getting an even ticket in the second draw is \( P(\text{second even}) = \frac{7}{16} \)

Since we draw without replacement, both events are dependent. The probability that both tickets are even is: \( P(\text{both even}) = \frac{8}{17} \times \frac{7}{16} = \frac{56}{272} = \frac{7}{34} \)

Exam Tip: For drawing without replacement, the second probability must reflect the reduced total and reduced count of favorable outcomes.

Question 2. Two marbles are drawn successively from a box containing 3 black and 4 white marbles. Find the probability that both the marbles are black if the first marble is not replaced before the second draw.
Answer: Given: A box has 3 black and 4 white marbles. Each trial is independent of the other.

The sample space is S = {1B, 2B, 3B, 1W, 2W, 3W, 4W}.

To find: The probability that both marbles drawn are black.

Let success mean the marble drawn is black, i.e., P = 3/7

The probability of success in the first trial is P₁(success) = 3/7

The probability of success in the second trial without replacement of the first draw is P₂(success) = 2/6 = 1/3

Hence, the probability that both marbles drawn are black, with each trial being independent, is given by P₁ × P₂ = 3/7 × 2/6 = 1/7

Exam Tip: For problems without replacement, always reduce both the numerator and denominator in the second probability by 1 to account for the first item removed from the sample.

Question 3. A card is drawn from a well-shuffled deck of 52 cards and without replacing this card, a second card is drawn. Find the probability that the first card is a club and the second card is a spade.
Answer: Given: A well-shuffled deck of 52 cards. Each draw is independent of the other.

To find: The probability that the first card drawn is a club and the second card is a spade.

Let success for the first trial mean getting a club.

The probability of success in the first trial is P₁(success) = 13/52

Let success for the second trial mean getting a spade.

The probability of success in the second trial without replacement of the first draw is P₂(success) = 13/51

Hence, the probability that the first card drawn is a club and the second card is a spade, with each trial being independent, is given by P₁ × P₂ = 13/52 × 13/51 = 169/2652 = 13/204

Exam Tip: Remember that in a standard deck, there are 13 cards of each suit. After removing one card, only 51 cards remain, but the count of the target suit stays the same since different suits are involved.

Question 4. There is a box containing 30 bulbs, of which 5 are defective. If two bulbs are chosen at random from the box in succession without replacing the first, what is the probability that both the bulbs are chosen are defective?
Answer: Given: A box contains 30 bulbs of which 5 are defective. Each trial is independent of the other.

To find: The probability that both bulbs chosen are defective.

Let success mean the bulb chosen is defective, i.e., P = 5/30 = 1/6

The probability of success in the first trial is P₁(success) = 5/30

The probability of success in the second trial without replacement of the first draw is P₂(success) = 4/29

Hence, the probability that both bulbs chosen are defective, with each trial being independent, is given by P₁ × P₂ = 5/30 × 4/29 = 20/870 = 2/87

Exam Tip: In successive sampling without replacement, always track both the reduced total and the reduced count of items matching your target condition.

Question 5. A bag contains 10 white and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that the first ball is white and the second is black?
Answer: Given: A bag contains 10 white and 15 black balls. Each trial is independent of the other.

To find: The probability that the first ball drawn is white and the second ball drawn is black.

Let success in the first draw mean getting a white ball.

The probability of success in the first trial is P₁(success) = 10/25 = 2/5

Let success in the second draw mean getting a black ball.

The probability of success in the second trial without replacement of the first draw is P₂(success) = 15/24 = 5/8

Hence, the probability that the first ball drawn is white and the second ball drawn is black, with each trial being independent, is given by P₁ × P₂ = 10/25 × 15/24 = 150/600 = 1/4

Exam Tip: When drawing different types of items in sequence, the second draw total is always one less, but the count of the second item type remains unchanged since you removed a different type first.

Question 6. An urn contains 5 white and 8 black balls. Two successive drawings of 3 balls at a time are made such that the balls drawn in the first draw are not replaced before the second draw. Find the probability that the first draw gives 3 white balls and the second draw gives 3 black balls.
Answer: Given: An urn contains 5 white and 8 black balls. Each trial is independent of the other.

To find: The probability that the first draw gives 3 white balls and the second draw gives 3 black balls.

Let success in the first draw mean getting 3 white balls.

The probability of success in the first trial is P₁(success) = C(5,3) / C(13,3) = 10/286 = 5/143

Let success in the second draw mean getting 3 black balls.

The probability of success in the second trial without replacement of the first draw is P₂(success) = C(8,3) / C(10,3) = 56/120 = 7/15

Hence, the probability that the first draw gives 3 white and the second draw gives 3 black balls, with each trial being independent, is given by P₁ × P₂ = 5/143 × 7/15 = 35/2145 = 7/429

Exam Tip: When selecting multiple items without replacement, use combinations. After the first draw, the urn has fewer total balls, which affects the denominator of the second probability.

Question 7. Let E₁ and E₂ be the events such that P(E₁) = 1/3 and P(E₂) = 3/5. Find: (i) P(E₁ ∪ E₂), when E₁ and E₂ are mutually exclusive. (ii) P(E₁ ∩ E₂), when E₁ and E₂ are independent.
Answer: Given: E₁ and E₂ are two events such that P(E₁) = 1/3 and P(E₂) = 3/5

To find: (i) P(E₁ ∪ E₂) when E₁ and E₂ are mutually exclusive.

We know that when two events are mutually exclusive, P(E₁ ∩ E₂) = 0

Hence, P(E₁ ∪ E₂) = P(E₁) + P(E₂)

= 1/3 + 3/5

= 5/15 + 9/15

= 14/15

Therefore, P(E₁ ∪ E₂) = 14/15 when E₁ and E₂ are mutually exclusive.

(ii) P(E₁ ∩ E₂) when E₁ and E₂ are independent.

We know that when E₁ and E₂ are independent,

P(E₁ ∩ E₂) = P(E₁) × P(E₂)

= 1/3 × 3/5

= 1/5

Therefore, P(E₁ ∩ E₂) = 1/5 when E₁ and E₂ are independent.

Exam Tip: Distinguish clearly between mutually exclusive events (no overlap, add probabilities) and independent events (occurrence of one does not affect the other, multiply probabilities).

Question 8. If E₁ and E₂ are the two events such that P(E₁) = 1/4, P(E₂) = 1/3 and P(E₁ ∪ E₂) = 1/2, show that E₁ and E₂ are independent events.
Answer: Given: E₁ and E₂ are two events such that P(E₁) = 1/4, P(E₂) = 1/3 and P(E₁ ∪ E₂) = 1/2

To show: E₁ and E₂ are independent events.

We know that P(E₁ ∩ E₂) = P(E₁) + P(E₂) - P(E₁ ∪ E₂)

= 1/4 + 1/3 - 1/2

= 3/12 + 4/12 - 6/12

= 1/12 ... (Equation 1)

Since the condition for two events to be independent is

P(E₁ ∩ E₂) = P(E₁) × P(E₂)

= 1/4 × 1/3

= 1/12 ... (Equation 2)

Since Equation 1 = Equation 2, E₁ and E₂ are independent events.

Hence proved.

Exam Tip: To verify independence, calculate the intersection probability both ways - using the union formula and using the product rule - and show they are equal.

Question 9. If E₁ and E₂ are independent events such that P(E₁) = 0.3 and P(E₂) = 0.4, find (i) P(E₁ ∩ E₂) (ii) P(E₁ ∪ E₂) (iii) P(\( \overline{E₁} \) ∩ \( \overline{E₂} \)) (iv) P(\( \overline{E₁} \) ∩ E₂)
Answer: Given: E₁ and E₂ are two independent events such that P(E₁) = 0.3 and P(E₂) = 0.4

To find: (i) P(E₁ ∩ E₂)

We know that when E₁ and E₂ are independent,

P(E₁ ∩ E₂) = P(E₁) × P(E₂)

= 0.3 × 0.4

= 0.12

Therefore, P(E₁ ∩ E₂) = 0.12 when E₁ and E₂ are independent.

(ii) P(E₁ ∪ E₂) when E₁ and E₂ are independent.

We know that P(E₁ ∪ E₂) = P(E₁) + P(E₂) - P(E₁ ∩ E₂)

= 0.3 + 0.4 - (0.3 × 0.4)

= 0.58

Therefore, P(E₁ ∪ E₂) = 0.58 when E₁ and E₂ are independent.

(iii) P(\( \overline{E₁} \) ∩ \( \overline{E₂} \)) = P(\( \overline{E₁} \)) × P(\( \overline{E₂} \))

Since P(E₁) = 0.3 and P(E₂) = 0.4

\( \Rightarrow \) P(\( \overline{E₁} \)) = 1 - P(E₁) = 0.7 and P(\( \overline{E₂} \)) = 1 - P(E₂) = 0.6

Since E₁ and E₂ are two independent events, \( \overline{E₁} \) and \( \overline{E₂} \) are also independent events.

Therefore, P(\( \overline{E₁} \) ∩ \( \overline{E₂} \)) = 0.7 × 0.6 = 0.42

(iv) P(\( \overline{E₁} \) ∩ E₂) = P(\( \overline{E₁} \)) × P(E₂)

= 0.7 × 0.4

= 0.28

Therefore, P(\( \overline{E₁} \) ∩ E₂) = 0.28

Exam Tip: Remember that if E₁ and E₂ are independent, then all combinations of their complements with the original events are also independent - use this to simplify calculations.

Question 10. Let A and B be the events such that P(A) = 1/2, P(B) = 7/12 and P(not A or not B) = 1/4. State whether A and B are (i) mutually exclusive (ii) independent.
Answer: Given: A and B are events such that P(A) = 1/2 and P(B) = 7/12 and P(not A or not B) = 1/4

To find: (i) If A and B are mutually exclusive

Since P(not A or not B) = 1/4, i.e., P(\( \overline{A} \) ∪ \( \overline{B} \)) = 1/4

We know that P(\( \overline{A} \) ∪ \( \overline{B} \)) = P(A ∩ B)' = 1 - P(A ∩ B) = 1/4

\( \Rightarrow \) P(A ∩ B) = 1 - 1/4 = 3/4 ... (Equation 1)

Since for two mutually exclusive events P(A ∩ B) = 0

But here P(A ∩ B) ≠ 0

Therefore, A and B are not mutually exclusive.

(ii) If A and B are independent

The condition for two events to be independent is P(E₁ ∩ E₂) = P(E₁) × P(E₂)

= 1/2 × 7/12

= 7/24 ... (Equation 2)

Since Equation 1 ≠ Equation 2

\( \Rightarrow \) A and B are not independent

Exam Tip: Use the complement rule strategically - converting "not A or not B" to its complement helps you find P(A ∩ B) directly, which then determines both mutual exclusivity and independence.

Question 11. Kamal and Vimal appeared for an interview for two vacancies. The probability of Kamal's selection is 1/3, and that of Vimal's selection is 1/5. Find the probability that only one of them will be selected.
Answer: Given: Let A denote the event 'Kamal is selected' and let B denote the event 'Vimal is selected'.

Therefore, P(A) = 1/3 and P(B) = 1/5

Also, A and B are independent, A and not B are independent, and not A and B are independent.

To find: The probability that only one of them will be selected.

Now, P(only one of them is selected) = P(A and not B or B and not A)

= P(A and not B) + P(B and not A)

= P(A ∩ \( \overline{B} \)) + P(B ∩ \( \overline{A} \))

= P(A) × P(\( \overline{B} \)) + P(B) × P(\( \overline{A} \))

= P(A) × [1 - P(B)] + P(B) × [1 - P(A)]

= 1/3 × [1 - 1/5] + 1/5 × [1 - 1/3]

= 1/3 × 4/5 + 1/5 × 2/3

= 4/15 + 2/15

= 6/15 = 2/5

Therefore, the probability that only one of them will be selected is 2/5

Exam Tip: For "only one" scenarios, add the probabilities of each person being selected alone (selected while the other is not) - this covers all cases where exactly one succeeds.

Question 12. Arun and Ved appeared for an interview for two vacancies. The probability of Arun's selection is 1/4, and that of Ved's rejection is 2/3. Find the probability that at least one of them will be selected.
Answer: Given: Let A denote the event 'Arun is selected' and let B denote the event 'Ved is selected'.

Therefore, P(A) = 1/4 and P(\( \overline{B} \)) = 2/3, so P(B) = 1/3 and P(\( \overline{A} \)) = 3/4

Also, A and B are independent, A and not B are independent, and not A and B are independent.

To find: The probability that at least one of them will get selected.

Now, P(at least one of them getting selected) = P(selecting only Arun) + P(selecting only Ved) + P(selecting both)

= P(A and not B) + P(B and not A) + P(A and B)

= P(A ∩ \( \overline{B} \)) + P(B ∩ \( \overline{A} \)) + P(A ∩ B)

= P(A) × P(\( \overline{B} \)) + P(B) × P(\( \overline{A} \)) + P(A) × P(B)

= 1/4 × 2/3 + 1/3 × 3/4 + 1/4 × 1/3

= 2/12 + 3/12 + 1/12

= 6/12 = 1/2

Therefore, the probability that at least one of them will get selected is 1/2

Exam Tip: For "at least one," break down into three mutually exclusive cases: first succeeds alone, second succeeds alone, and both succeed - then add their probabilities.

Question 13. A and B appear for an interview for two vacancies in the same post. The probability of A's selection is 1/6 and that of B's selection is 1/4. Find the probability that (i) both of them are selected (ii) only one of them is selected (iii) none is selected (iv) at least one of them is selected.
Answer: Given: A and B appear for an interview. Then P(A) = 1/6 and P(B) = 1/4, so P(\( \overline{A} \)) = 5/6 and P(\( \overline{B} \)) = 3/4

Also, A and B are independent, A and not B are independent, and not A and B are independent.

To find: (i) The probability that both of them are selected.

We know that P(both of them are selected) = P(A ∩ B) = P(A) × P(B)

= 1/6 × 1/4

= 1/24

Therefore, the probability that both of them are selected is 1/24

(ii) P(only one of them is selected) = P(A and not B or B and not A)

= P(A and not B) + P(B and not A)

= P(A ∩ \( \overline{B} \)) + P(B ∩ \( \overline{A} \))

= P(A) × P(\( \overline{B} \)) + P(B) × P(\( \overline{A} \))

= [1/6 × 3/4] + [1/4 × 5/6]

= 3/24 + 5/24

= 8/24 = 1/3

Therefore, the probability that only one of them is selected is 1/3

(iii) None is selected

We know that P(none is selected) = P(\( \overline{A} \) ∩ \( \overline{B} \))

= P(\( \overline{A} \)) × P(\( \overline{B} \))

= 5/6 × 3/4

= 15/24 = 5/8

Therefore, the probability that none is selected is 5/8

(iv) At least one of them is selected

Now, P(at least one of them is selected) = P(selecting only A) + P(selecting only B) + P(selecting both)

= P(A and not B) + P(B and not A) + P(A and B)

= P(A ∩ \( \overline{B} \)) + P(B ∩ \( \overline{A} \)) + P(A ∩ B)

= P(A) × P(\( \overline{B} \)) + P(B) × P(\( \overline{A} \)) + P(A) × P(B)

= [1/6 × 3/4] + [1/4 × 5/6] + [1/6 × 1/4]

= 3/24 + 5/24 + 1/24

= 9/24 = 3/8

Therefore, the probability that at least one of them is selected is 3/8

Exam Tip: For exhaustive probability questions with multiple scenarios, verify your answers by checking that all probabilities sum to 1 (both selected + only one + none = 1).

Question 14. Given the probability that A can solve a problem is 2/3, and the probability that B can solve the same problem is 3/5, find the probability that (i) at least one of A and B will solve the problem (ii) none of the two will solve the problem.
Answer: Given: The probability that A and B can solve the same problem is given, i.e., P(A) = 2/3 and P(B) = 3/5, so P(\( \overline{A} \)) = 1/3 and P(\( \overline{B} \)) = 2/5

Also, A and B are independent, and not A and not B are independent.

To find: (i) At least one of A and B will solve the problem

Now, P(at least one of them will solve the problem) = 1 - P(both are unable to solve)

= 1 - P(\( \overline{A} \) ∩ \( \overline{B} \))

= 1 - P(\( \overline{A} \)) × P(\( \overline{B} \))

= 1 - [1/3 × 2/5]

= 1 - 2/15

= 13/15

Therefore, at least one of A and B will solve the problem with probability 13/15

(ii) None of the two will solve the problem

Now, P(none of the two will solve the problem) = P(\( \overline{A} \) ∩ \( \overline{B} \))

= P(\( \overline{A} \)) × P(\( \overline{B} \))

= 1/3 × 2/5

= 2/15

Therefore, the probability that none of the two will solve the problem is 2/15

Exam Tip: For "at least one" problems, use the complement method: P(at least one) = 1 - P(none), which is often faster than adding all individual scenarios.

Question 15. A problem is given to three students whose chances of solving it are 1/4, 1/5 and 1/6, respectively. Find the probability that the problem is solved.
Answer: Given: Let A, B, and C be three students whose chances of solving a problem are P(A) = 1/4, P(B) = 1/5, and P(C) = 1/6

P(\( \overline{A} \)) = 3/4, P(\( \overline{B} \)) = 4/5, and P(\( \overline{C} \)) = 5/6

To find: The probability that the problem is solved.

Here, P(the problem is solved) = 1 - P(the problem is not solved)

= 1 - P(\( \overline{A} \) ∩ \( \overline{B} \) ∩ \( \overline{C} \))

= 1 - [P(\( \overline{A} \)) × P(\( \overline{B} \)) × P(\( \overline{C} \))]

= 1 - [3/4 × 4/5 × 5/6]

= 1 - 60/120

= 1 - 1/2

= 1/2

Therefore, the probability that the problem is solved is 1/2

Exam Tip: When multiple independent agents attempt a task, always use the complement: P(at least one succeeds) = 1 - P(all fail), as this avoids dealing with overlapping success scenarios.

Question 16. The probabilities of A, B, C solving a problem are 1/3, 1/4 and 1/6, respectively. If all the three try to solve the problem simultaneously, find the probability that exactly one of them will solve it.
Answer: Given: Let A, B, and C be three students whose chances of solving a problem are P(A) = 1/3, P(B) = 1/4, and P(C) = 1/6

P(\( \overline{A} \)) = 2/3, P(\( \overline{B} \)) = 3/4, and P(\( \overline{C} \)) = 5/6

To find: The probability that exactly one of them will solve it.

Now, P(exactly one of them will solve it) = P(A and not B and not C) + P(B and not A and not C) + P(C and not A and not B)

= P(A ∩ \( \overline{B} \) ∩ \( \overline{C} \)) + P(B ∩ \( \overline{A} \) ∩ \( \overline{C} \)) + P(C ∩ \( \overline{A} \) ∩ \( \overline{B} \))

= P(A) × P(\( \overline{B} \)) × P(\( \overline{C} \)) + P(B) × P(\( \overline{A} \)) × P(\( \overline{C} \)) + P(C) × P(\( \overline{A} \)) × P(\( \overline{B} \))

= [1/3 × 3/4 × 5/6] + [1/4 × 2/3 × 5/6] + [1/6 × 2/3 × 3/4]

= 15/72 + 10/72 + 6/72

= 31/72

Therefore, the probability that exactly one of them will solve the problem is 31/72

Exam Tip: For "exactly one" scenarios, identify all three mutually exclusive cases (each person solving alone) and add their individual probabilities carefully, multiplying the complementary probabilities for those who do not solve.

Question 17. A can hit a target 4 times in 5 shots, B can hit 3 times in 4 shots, and C can hit 2 times in 3 shots. Calculate the probability that (i) A, B and C all hit the target (ii) B and C hit and A does not hit the target.
Answer: Given: Let A, B, and C represent chances of hitting a target. So P(A) = 4/5, P(B) = 3/4, and P(C) = 2/3

P(\( \overline{A} \)) = 1/5, P(\( \overline{B} \)) = 1/4, and P(\( \overline{C} \)) = 1/3

To find: (i) The probability that A, B, and C all hit the target.

Now, P(all hitting the target) = P(A ∩ B ∩ C)

= P(A) × P(B) × P(C)

= 4/5 × 3/4 × 2/3

= 24/60 = 2/5

Hence, the probability that A, B, and C all hit the target is 2/5

(ii) B and C hit and A does not hit the target

Here, P(B and C hit and not A) = P(B ∩ C ∩ \( \overline{A} \))

= P(B) × P(C) × P(\( \overline{A} \))

= 3/4 × 2/3 × 1/5

= 6/60 = 1/10

Hence, the probability that B and C hit and A does not hit the target is 1/10

Exam Tip: When multiple people engage in independent attempts, multiply their individual success (or failure) probabilities in the required combination to find the desired outcome.

Question 18. Neelam has offered physics, chemistry and mathematics in Class XII. She estimates that her probabilities of receiving a grade A in these courses are 0.2, 0.3 and 0.9 respectively. Find the probabilities that Neelam receives (i) all A grades (ii) no A grade (iii) exactly 2 A grades.
Answer: Given: Let A, B, and C represent the subjects physics, chemistry, and mathematics respectively. The probability of Neelam getting an A grade in these three subjects is P(A) = 0.2, P(B) = 0.3, and P(C) = 0.9

P(\( \overline{A} \)) = 0.8, P(\( \overline{B} \)) = 0.7, and P(\( \overline{C} \)) = 0.1

To find: (i) The probability that Neelam gets all A grades

Here, P(getting all A grades) = P(A ∩ B ∩ C)

= P(A) × P(B) × P(C)

= 0.2 × 0.3 × 0.9

= 0.054

Therefore, the probability that Neelam gets all A grades is 0.054

(ii) No A grade

Here, P(getting no A grade) = P(\( \overline{A} \) ∩ \( \overline{B} \) ∩ \( \overline{C} \))

= P(\( \overline{A} \)) × P(\( \overline{B} \)) × P(\( \overline{C} \))

= 0.8 × 0.7 × 0.1

= 0.056

Therefore, the probability that Neelam gets no A grade is 0.056

(iii) Exactly 2 A grades

P(getting exactly 2 A grades) = P(A and B and not C) + P(B and C and not A) + P(C and A and not B)

= P(A ∩ B ∩ \( \overline{C} \)) + P(B ∩ C ∩ \( \overline{A} \)) + P(C ∩ A ∩ \( \overline{B} \))

= P(A) × P(B) × P(\( \overline{C} \)) + P(B) × P(C) × P(\( \overline{A} \)) + P(C) × P(A) × P(\( \overline{B} \))

= [0.2 × 0.3 × 0.1] + [0.3 × 0.9 × 0.8] + [0.9 × 0.2 × 0.7]

= 0.006 + 0.216 + 0.126

= 0.348

Therefore, the probability that Neelam gets exactly 2 A grades is 0.348

Exam Tip: For "exactly k successes" among n independent trials, enumerate all possible combinations where k succeed and the rest fail, then add their probabilities.

Question 19. An article manufactured by a company consists of two parts X and Y. In the process of manufacture of part X, 8 out of 100 parts may be defective. Similarly, 5 out of 100 parts of Y may be defective. Calculate the probability that the assembled product will not be defective.
Answer: Given: X and Y are the two parts of a company that manufactures an article.

The probability of the parts being defective is given, i.e., P(X) = 8/100 and P(Y) = 5/100, so P(\( \overline{X} \)) = 92/100 and P(\( \overline{Y} \)) = 95/100

To find: The probability that the assembled product will not be defective.

Here, P(product assembled will not be defective) = 1 - P(product assembled to be defective)

= 1 - [P(X and not Y) + P(Y and not X) + P(both)]

= 1 - [P(X ∩ \( \overline{Y} \)) + P(Y ∩ \( \overline{X} \)) + P(X ∩ Y)]

= 1 - [P(X) × P(\( \overline{Y} \)) + P(Y) × P(\( \overline{X} \)) + P(X) × P(Y)]

= 1 - [8/100 × 95/100 + 5/100 × 92/100 + 8/100 × 5/100]

= 1 - [760/10000 + 460/10000 + 40/10000]

= 1 - 1260/10000

= 8740/10000 = 437/500

Therefore, the probability that the assembled product will not be defective is 437/500

Exam Tip: For assembled products where at least one part failing causes failure, use the complement: P(product works) = 1 - P(at least one part fails).

Question 20. A town has two fire-extinguishing engines, functioning independently. The probability of availability of each engine when needed is 0.95. What is the probability that (i) neither of them is available when needed? (ii) an engine is available when needed?
Answer: Given: Let A and B be two fire-extinguishing engines. The probability of availability of each of the two fire-extinguishing engines is given, i.e., P(A) = 0.95 and P(B) = 0.95, so P(\( \overline{A} \)) = 0.05 and P(\( \overline{B} \)) = 0.05

To find: (i) The probability that neither of them is available when needed

Here, P(not A and not B) = P(\( \overline{A} \) ∩ \( \overline{B} \))

= P(\( \overline{A} \)) × P(\( \overline{B} \))

= 0.05 × 0.05

= 0.0025 = 1/400

Therefore, the probability that neither of them is available when needed is 1/400

(ii) An engine is available when needed

Here, P(A and not B or B and not A) = P(A ∩ \( \overline{B} \)) + P(B ∩ \( \overline{A} \))

= P(A) × P(\( \overline{B} \)) + P(B) × P(\( \overline{A} \))

= (0.95 × 0.05) + (0.95 × 0.05)

= 0.0475 + 0.0475

= 0.095 = 19/200

Therefore, the probability that an engine is available when needed is 19/200

Exam Tip: For "exactly one available" among identical-probability systems, you can multiply one availability and one unavailability, then double the result.

Question 21. A machine operates only when all of its three components function. The probabilities of the failures of the first, second and third components are 0.14, 0.10 and 0.05, respectively. What is the probability that the machine will fail?
Answer: Given: Let A, B, and C be the three components of a machine which works only if all three components function. The probabilities of the failures of A, B, and C respectively are P(A) = 0.14, P(B) = 0.10, and P(C) = 0.05

P(\( \overline{A} \)) = 0.86, P(\( \overline{B} \)) = 0.90, and P(\( \overline{C} \)) = 0.95

To find: The probability that the machine will fail.

Here, P(the machine will fail) = 1 - P(the machine will function)

= 1 - P(all three components are working)

= 1 - P(\( \overline{A} \) ∩ \( \overline{B} \) ∩ \( \overline{C} \))

= 1 - [P(\( \overline{A} \)) × P(\( \overline{B} \)) × P(\( \overline{C} \))]

= 1 - [0.86 × 0.90 × 0.95]

= 1 - 0.7353

= 0.2647

Therefore, the probability that the machine will fail is 0.2647

Exam Tip: When all components must work for the system to work, the system fails if any single component fails - use P(system fails) = 1 - P(all components work).

Question 22. An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shots are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that at least one shot hits the plane?
Answer: Given: Let A, B, C, and D be the first, second, third, and fourth shots whose probability of hitting the plane is given, i.e., P(A) = 0.4, P(B) = 0.3, P(C) = 0.2, and P(D) = 0.1 respectively

P(\( \overline{A} \)) = 0.6, P(\( \overline{B} \)) = 0.7, P(\( \overline{C} \)) = 0.8, and P(\( \overline{D} \)) = 0.9

To find: The probability that at least one shot hits the plane.

Here, P(at least one shot hits the plane) = 1 - P(none of the shots hit the plane)

= 1 - P(\( \overline{A} \) ∩ \( \overline{B} \) ∩ \( \overline{C} \) ∩ \( \overline{D} \))

= 1 - [P(\( \overline{A} \)) × P(\( \overline{B} \)) × P(\( \overline{C} \)) × P(\( \overline{D} \))]

= 1 - [0.6 × 0.7 × 0.8 × 0.9]

= 1 - 0.3024

= 0.6976

Therefore, the probability that at least one shot hits the plane is 0.6976

Exam Tip: The complement method P(at least one hits) = 1 - P(all miss) is far more efficient than summing all individual hit scenarios across multiple attempts.

Question 23. Let S₁ and S₂ be the two switches and let their probabilities of working be given by P(S₁) = 4/5 and P(S₂) = 9/10. Find the probability that the current flows from the terminal A to terminal B when S₁ and S₂ are installed in series, shown as follows:
Answer: Given: S₁ and S₂ are two switches whose probabilities of working are given by P(S₁) = 4/5 and P(S₂) = 9/10

To find: The probability that the current flows from terminal A to terminal B when S₁ and S₂ are connected in series.

Now, since the current in series flows from end to end, the flow of current from terminal A to terminal B is given by

P(S₁ ∩ S₂) = P(S₁) × P(S₂)

= 4/5 × 9/10

= 36/50 = 18/25

Therefore, the probability that the current flows from terminal A to terminal B when S₁ and S₂ are connected in series is 18/25

Exam Tip: In a series circuit, current flows only if all switches work - multiply their individual probabilities. In a parallel circuit, current flows if at least one switch works - use the union formula.

Question 24. Let S₁ and S₂ be two the switches and let their probabilities of working be given by P(S₁) = 2/3 and P(S₂) = 3/4. Find the probability that the current flows from terminal A to terminal B, when S₁ and S₂ are installed in parallel, as shown below:
Answer: Given: S₁ and S₂ are two switches whose probabilities of working are given by P(S₁) = 2/3 and P(S₂) = 3/4

To find: The probability that the current flows from terminal A to terminal B when S₁ and S₂ are connected in parallel.

Now, since current in parallel flows in two or more paths and hence the sum of currents through each path equals the total current flowing from the source, the flow of current from terminal A to terminal B in a parallel circuit is given by

P(S₁ ∪ S₂) = P(S₁) + P(S₂) - P(S₁ ∩ S₂)

= P(S₁) + P(S₂) - [P(S₁) × P(S₂)]

= 2/3 + 3/4 - [2/3 × 3/4]

= 8/12 + 9/12 - 6/12

= 11/12

Therefore, the probability that the current flows from terminal A to terminal B when S₁ and S₂ are connected in parallel is 11/12

Exam Tip: For parallel circuits, use the union formula because current can flow through either switch (or both) - this gives you P(at least one works).

Question 25. A coin is tossed. If a head comes up, a die is thrown, but if a tail comes up, the coin is tossed again. Find the probability of obtaining (i) two tails (ii) a head and the number 6 (iii) a head and an even number.
Answer: Given: Let H be head and T be tails, where 1, 2, 3, 4, 5, 6 are the numbers on the die which are thrown when a head comes up, or else the coin is tossed again if it shows tail.

According to the question, sample space S = {(TH), (TT), (H1), (H2), (H3), (H4), (H5), (H6)}

To find: (i) The probability of obtaining two tails

From the sample space, it is clear that the probability of obtaining two tails is 1/8, i.e., {TT} with total number of elements in sample space as 8.

Exam Tip: When analyzing compound experiments with conditional steps, explicitly list the complete sample space to identify all possible outcomes clearly.

Question 1. What is the probability of obtaining a head and the number 6?
Answer: By examining the sample space, we can see that getting a head along with the number 6 happens in just one case - {H6}. Since there are 8 total possible outcomes in the sample space, the probability is \( \frac{1}{8} \).
In simple words: Out of all 8 possible results when you flip a coin and roll a die at the same time, only one result gives you both a head and a 6. So the chance of this happening is 1 out of 8.

Exam Tip: Always identify the favorable outcome(s) clearly and divide by the total number of outcomes in the sample space - this is the fundamental definition of probability.

Question 2. What is the probability of obtaining a head and an even number?
Answer: Looking at the sample space, the outcomes where we get a head paired with an even number are {H2, H4, H6}. This gives us 3 favorable outcomes. With 8 total possible outcomes in the sample space, the probability is \( \frac{3}{8} \).
In simple words: When you flip a coin and roll a die together, there are three ways to get a head and an even number - you can roll a 2, 4, or 6 with the head. So out of 8 total chances, 3 of them work. The probability is 3 out of 8.

Exam Tip: Count all outcomes that satisfy both conditions (head AND even number) - don't forget to include all even numbers on the die (2, 4, and 6).