Maharashtra Board Class 11 Maths Part 2 Chapter 7 Probability 7.4 Solutions

Question 1. Two dice are thrown simultaneously, if at least one of the dice shows a number 5, what is the probability that sum of the numbers on two dice is 9?
Answer: Solution: When two dice are thrown simultaneously, the sample space is S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} ∴ n(S) = 36 Let A be the event that at least one die shows number 5. ∴ A = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5)} ∴ n(A) = 11 ∴ \(P(A) = \frac{n(A)}{n(S)} = \frac{11}{36}\) Let B be the event that sum of the numbers on two dice is 9. ∴ B = {(3, 6), (4, 5), (5, 4), (6, 3)} Also, A ∩ B = {(4, 5), (5, 4)} ∴ n(A ∩ B) = 2 ∴ \(P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{36}\) ∴ Probability of sum of numbers on two dice is 9, given that one dice shows number 5, is given by \(P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{2}{36}}{\frac{11}{36}} = \frac{2}{11}\)
In simple words: This problem involves conditional probability, where we find the probability of one event (sum is 9) happening given that another event (at least one die shows 5) has already occurred when two dice are rolled.

🎯 Exam Tip: Clearly define the sample space (S), and the events (A and B) along with their intersections to correctly calculate conditional probabilities.

Question 2. A pair of dice is thrown. If sum of the numbers is an even number, what is the probability that it is a perfect square?
Answer: Solution: When two dice are thrown simultaneously, the sample space is S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} ∴ n(S) = 36 Let A be the event that sum of the numbers is an even number. ∴ A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)} ∴ n(A) = 18 ∴ \(P(A) = \frac{n(A)}{n(S)} = \frac{18}{36}\) Let B be the event that sum of outcomes is a perfect square. ∴ B = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)} Also, A ∩ B= {(1, 3), (2, 2), (3, 1)} ∴ n(A ∩ B) = 3 ∴ \(P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{3}{36}\) ∴ Probability of sum of the numbers is a perfect square, given that sum of numbers is an even number, is given by \(P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{3}{36}}{\frac{18}{36}} = \frac{3}{18} = \frac{1}{6}\)
In simple words: This question asks for the conditional probability of getting a perfect square sum, given that the sum of the numbers on two dice is an even number. We list all possible outcomes, then identify outcomes for each event and their intersection.

🎯 Exam Tip: Always list all elements of the sample space and relevant events clearly to avoid counting errors, especially for conditional probability problems.

Question 3. A box contains 11 tickets numbered from 1 to 11. Two tickets are drawn at random with replacement. If the sum is even, find the probability that both the numbers are odd.
Answer: Solution: Two tickets can be drawn from 11 tickets with replacement in \(11 \times 11 = 121\) ways. ∴ n(S) = 121 Let A be the event that the sum of two numbers is even. The event A occurs, if either both the tickets with odd numbers or both the tickets with even numbers are drawn. There are 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10) from 1 to 11. ∴ n(A) = \(6 \times 6 + 5 \times 5\) = 36 + 25 = 61 ∴ \(P(A) = \frac{n(A)}{n(S)} = \frac{61}{121}\) Let B be the event that the numbers tickets drawn are odd ∴ n(B) = \(6 \times 6 = 36\) ∴ \(P(B) = \frac{n(B)}{n(S)} = \frac{36}{121}\) Since 6 odd numbers are common between A and B. ∴ n(A ∩ B) = \(6 \times 6 = 36\) ∴ \(P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{36}{121}\) ∴ Probability of both the numbers are odd, given that sum is even, is given by \(P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{36}{121}}{\frac{61}{121}} = \frac{36}{61}\)
In simple words: This problem asks for the probability that both drawn tickets are odd, given that their sum is even, with replacement. We calculate probabilities for the sum being even, both tickets being odd, and their intersection.

🎯 Exam Tip: When drawing with replacement, the total number of outcomes for successive draws is found by multiplying the number of options for each draw. Be careful with definitions of events A and B.

Question 4. A card is drawn from a well-shuffled pack of 52 cards. Consider two events A and B as A: a club card is drawn. B: an ace card is drawn. Determine whether events A and B are independent or not.
Answer: Solution: One card can be drawn out of 52 cards in \(^{52}C_1\) ways. ∴ n(S) = \(^{52}C_1\) Let A be the event that a club card is drawn. 1 club card out of 13 club cards can be drawn in \(^{13}C_1\) ways. ∴ n(A) = \(^{13}C_1\) ∴ \(P(A) = \frac{n(A)}{n(S)} = \frac{^{13}C_1}{^{52}C_1}\) Let B be the event that an ace card is drawn. An ace card out of 4 aces can be drawn in \(^{4}C_1\) ways. ∴ n(B) = \(^{4}C_1\) ∴ \(P(B) = \frac{n(B)}{n(S)} = \frac{^{4}C_1}{^{52}C_1}\) Since 1 card is common between A and B ∴ n(A ∩ B) = \(^{1}C_1\) ∴ \(P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{^{1}C_1}{^{52}C_1} = \frac{1}{52}\) .........(i) ∴ \(P(A) \times P(B) = \frac{^{13}C_1}{^{52}C_1} \times \frac{^{4}C_1}{^{52}C_1} = \frac{13 \times 4}{52 \times 52} = \frac{52}{52 \times 52} = \frac{1}{52}\) .........(ii) From (i) and (ii), we get P(A ∩ B) = P(A) × P(B) ∴ A and B are independent events.
In simple words: To determine if events A (drawing a club) and B (drawing an ace) are independent, we calculate the probabilities of A, B, and their intersection. If \(P(A \cap B) = P(A) \times P(B)\), then the events are independent.

🎯 Exam Tip: Remember that in a standard deck, there is only one Ace of Clubs. This overlap is crucial for calculating \(P(A \cap B)\) correctly to test for independence.

Question 5. A problem in statistics is given to three students A, B, and C. Their chances of solving the problem are 1/3, 1/4, and 1/5 respectively. If all of them try independently, what is the probability that, (i) problem is solved? (ii) problem is not solved? (iii) exactly two students solve the problem?
Answer: Solution: Let A be the event that student A can solve the problem. B be the event that student B can solve the problem. C be the event that student C can solve problem. ∴ \(P(A) = \frac{1}{3}\), \(P(B) = \frac{1}{4}\), \(P(C) = \frac{1}{5}\) ∴ \(P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}\) ∴ \(P(B') = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4}\) ∴ \(P(C') = 1 - P(C) = 1 - \frac{1}{5} = \frac{4}{5}\) Since A, B, C are independent events ∴ A', B', C' are also independent events (i) Let X be the event that problem is solved. Problem can be solved if at least one of the three students solves the problem. P(X) = P(at least one student solves the problem) = 1 – P(no student solved problem) = 1 – P(A' ∩ B' ∩ C') = 1 – P(A') P(B') P(C') = \(1 - \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}\) = \(1 - \frac{2}{5}\) = \(\frac{3}{5}\) (ii) Let Y be the event that problem is not solved ∴ P(Y) = P(A' ∩ Β' ∩ C') = P(A') P(Β') P(C') = \(\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}\) = \(\frac{2}{5}\) (iii) Let Z be the event that exactly two students solve the problem. ∴ P(Z) = P(A ∩ B ∩ C') U P(A ∩ Β' ∩ C) U P(A' ∩ B ∩ C) = P(A) . P(B) . P(C') + P(A) . P(B') . P(C) + P(A') . P(B) . P(C) = \((\frac{1}{3} \times \frac{1}{4} \times \frac{4}{5}) + (\frac{1}{3} \times \frac{3}{4} \times \frac{1}{5}) + (\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5})\) = \(\frac{4}{60} + \frac{3}{60} + \frac{2}{60}\) = \(\frac{9}{60}\) = \(\frac{3}{20}\)
In simple words: This problem explores probabilities related to three independent events. We calculate the chance of at least one student solving the problem, none solving it, and exactly two solving it by using the properties of independent events and their complements.

🎯 Exam Tip: For "at least one" problems, it's often simpler to calculate 1 minus the probability of "none" of the events occurring. For "exactly two", consider all combinations where two specific events occur and one does not.

Question 6. The probability that a 50-year old man will be alive till age 60 is 0.83 and the probability that a 45-year old woman will be alive till age 55 is 0.97. What is the probability that a man whose age is 50 and his wife whose age is 45 will both be alive for the next 10 years?
Answer: Solution: Let A be the event that man will be alive at 60. ∴ P(A) = 0.83 Let B be the event that a woman will be alive at 55. ∴ P(B) = 0.97 A ∩ B = Event that both will be alive. Also, A and B are independent events ∴ P(both man and his wife will be alive) = P(A ∩ B) = P(A) . P(B) = 0.83 × 0.97 = 0.8051
In simple words: This problem requires finding the probability of two independent events occurring simultaneously: the man being alive at 60 and the woman being alive at 55. Since the events are independent, we simply multiply their individual probabilities.

🎯 Exam Tip: For independent events, the probability of both events occurring is the product of their individual probabilities. Always state the independence assumption clearly if it's not explicitly given.

Question 7. In an examination, 30% of the students have failed in subject I, 20% of the students have failed in subject II and 10% have failed in both subjects I and subject II. A student is selected at random, what is the probability that the student (i) has failed in the subject I, if it is known that he is failed in subject II? (ii) has failed in at least one subject? (iii) has failed in exactly one subject?
Answer: Solution: Let A be the event that the student failed in Subject I B be the event that the student failed in Subject II Then \(P(A) = 30\% = \frac{30}{100}\) \(P(B) = 20\% = \frac{20}{100}\) and \(P(A \cap B) = 10\% = \frac{10}{100}\) (i) P (student failed in Subject I, given that he has failed in Subject II) = P(A/B) \( = \frac{P(A \cap B)}{P(B)} = \frac{\frac{10}{100}}{\frac{20}{100}} = \frac{10}{20} = \frac{1}{2}\) (ii) P(student failed in at least one subject) = P(A U B) = P(A) + P(B) – P(A ∩ B) \( = \frac{30}{100} + \frac{20}{100} - \frac{10}{100}\) = 0.40 (iii) P(student failed in exactly one subject) = P(A) + P(B) – 2P(A ∩ B) \( = \frac{30}{100} + \frac{20}{100} - 2(\frac{10}{100})\) = 0.30
In simple words: This problem involves probabilities of student failures in two subjects. We calculate conditional probability for failing Subject I given failure in Subject II, probability of failing at least one subject, and probability of failing exactly one subject using the formulas for conditional probability and union of events.

🎯 Exam Tip: Pay close attention to the wording: "at least one" implies the union formula, while "exactly one" means the sum of probabilities of (A and not B) plus (B and not A), which simplifies to P(A) + P(B) - 2P(A ∩ B).

Question 8. One-shot is fired from each of the three guns. Let A, B, and C denote the events that the target is hit by the first, second and third gun respectively. Assuming that A, B, and C are independent events and that P(A) = 0.5, P(B) = 0.6, and P(C) = 0.8, then find the probability that at least one hit is registered.
Answer: Solution: A be the event that first gun hits the target B be the event that second gun hits the target C be the event that third gun hits the target P(A) = 0.5, P(B) = 0.6, P(C) = 0.8 ∴ P(A') = 1 – P(A) = 1 – 0.5 = 0.5 ∴ P(B') = 1 – P(B) = 1 – 0.6 = 0.4 ∴ P(C') = 1 – P(C) = 1 – 0.8 = 0.2 Now A, B, C are independent events ∴ A', B', C are also independent events. ∴ P (at least one hit is registered) = 1 – P(no hit is registered) = 1 – P(A' ∩ B' ∩ C') = 1 – P(A') P(B') P(C') = 1 – (0.5) (0.4) (0.2) = 1 - 0.04 = 0.96
In simple words: To find the probability of at least one hit, we calculate the complement: 1 minus the probability of no hits from any of the independent guns. This involves finding the probabilities of each gun missing and multiplying them together.

🎯 Exam Tip: The "at least one" probability calculation using the complement rule (\(1 - P(\text{none})\)) is highly efficient for independent events, simplifying the process of considering multiple combinations.

Question 9. A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that (i) first is white and second is black? (ii) one is white and the other is black?
Answer: Solution: Total number of balls = 10 + 15 = 25 Let S be an event that two balls are drawn at random without replacement in succession ∴ n(S) = \(^{25}C_1 \times ^{24}C_1 = 25 \times 24 = 600\) (Corrected n(S) calculation for successive draws, though the solutions use different methods in places. Stick to the logic presented in the solution for consistency) (i) Let A be the event that the first ball is white and the second is black. First white ball can be drawn from 10 white balls in \(^{10}C_1\) ways and second black ball can be drawn from 15 black balls in \(^{15}C_1\) ways. ∴ n(A) = \(^{10}C_1 \times ^{15}C_1 = 10 \times 15 = 150\) ∴ \(P(A) = \frac{n(A)}{n(S)} = \frac{10 \times 15}{25 \times 24} = \frac{150}{600} = \frac{1}{4}\) (ii) Let B be the event that one is white and the other is black. This can happen if: - First is white, second is black (WB) - First is black, second is white (BW) Number of ways for WB = \(^{10}C_1 \times ^{15}C_1 = 10 \times 15 = 150\) Number of ways for BW = \(^{15}C_1 \times ^{10}C_1 = 15 \times 10 = 150\) ∴ n(B) = \(10C_1 \cdot 15C_1 + 15C_1 \cdot 10C_1 = 150 + 150 = 300\) ∴ \(P(B) = \frac{n(B)}{n(S)} = \frac{10 \times 15 + 15 \times 10}{25 \times 24} = \frac{150 + 150}{600} = \frac{300}{600} = \frac{1}{2}\)
In simple words: When drawing balls without replacement, the probability changes for the second draw. For "first white, second black," we multiply the probability of drawing a white first by the probability of drawing a black second from the remaining balls. For "one white, one black," we consider both orders: white then black, and black then white.

🎯 Exam Tip: In "without replacement" problems, remember that both the total number of items and the number of specific items decrease after each draw, impacting subsequent probabilities. Always consider all possible favorable sequences for "one of each" scenarios.

Question 10. An urn contains 4 black, 5 white, and 6 red balls. Two balls are drawn one after the other without replacement, what is the probability that at least one ball is black?
Answer: Solution: Total number of balls in the urn = 4 + 5 + 6 = 15 Two balls can be drawn without replacement in \(^{15}C_2 = \frac{15 \times 14}{1 \times 2} = 105\) ways ∴ n(S) = 105 Let A be the event that at least one ball is black i.e., 1 black and 1 non-black or 2 black and 0 non-black. 1 black ball can be drawn out of 4 black balls in \(^{4}C_1 = 4\) ways and 1 non-black ball can be drawn out of remaining 11 non-black balls in \(^{11}C_1 = 11\) ways ∴ 1 black and 1 non black ball can be drawn in \(4 \times 11 = 44\) ways Also, 2 black balls can be drawn from 4 black balls in \(^{4}C_2 = \frac{4 \times 3}{1 \times 2} = 6\) ways ∴ n(A) = 44 + 6 = 50 ∴ Required probability = \(P(A) = \frac{n(A)}{n(S)} = \frac{50}{105} = \frac{10}{21}\) Alternate Solution: Total number of balls = 15 Required probability = 1 – P(neither of two balls is black) Balls are drawn without replacement Probability of first non-black ball drawn = \(\frac{11}{15}\) Probability of second non-black ball drawn = \(\frac{10}{14}\) Probability of neither of two balls is black = \(\frac{11}{15} \times \frac{10}{14} = \frac{11}{21}\) Required probability = \(1 - \frac{11}{21} = \frac{10}{21}\)
In simple words: To find the probability of at least one black ball being drawn without replacement, you can either sum the probabilities of drawing exactly one black ball and drawing two black balls, or more efficiently, calculate 1 minus the probability of drawing no black balls at all.

🎯 Exam Tip: The complement rule (1 - P(none)) is often the easiest and most reliable method for "at least one" probability questions, especially when direct calculation involves multiple mutually exclusive scenarios.

Question 11. Two balls are drawn from an urn containing 5 green, 3 blue, 7 yellow balls one by one without replacement. What is the probability that at least one ball is blue?
Answer: Solution: Total number of balls in the urn = 5 + 3 + 7 = 15 Out of these 12 are non-blue balls. Two balls can be drawn from 15 balls without replacement in \(^{15}C_2\) \( = \frac{15 \times 14}{1 \times 2}\) = 105 ways. ∴ n(S) = 105 Let A be the event that at least one ball is blue, i.e., 1 blue and other non-blue or both are blue. ∴ n(A) = \(^{3}C_1 \times ^{12}C_1 + ^{3}C_2\) = \(3 \times 12 + 3\) = 36 + 3 = 39 ∴ \(P(A) = \frac{n(A)}{n(S)} = \frac{39}{105} = \frac{13}{35}\) Alternate solution: Total number of balls in the urn = 15 Required probability = 1 – P(neither of two balls is blue) Balls are drawn one by one without replacement. Probability of first non-blue ball drawn = \(\frac{12}{15}\) Probability of second non-blue ball drawn = \(\frac{11}{14}\) Probability of neither of two ball is blue = \(\frac{12}{15} \times \frac{11}{14} = \frac{22}{35}\) ∴ Required probability = \(1 - \frac{22}{35} = \frac{13}{35}\)
In simple words: This problem asks for the probability of drawing at least one blue ball when two are drawn without replacement. This can be calculated directly by considering the cases of one blue and one non-blue, or two blue, or by using the complement rule (1 minus the probability of drawing no blue balls).

🎯 Exam Tip: For "at least one" scenarios, the complement method is generally more straightforward. Ensure accurate calculation of combinations for the sample space and individual events when not using the complement approach.

Question 12. A bag contains 4 blue and 5 green balls. Another bag contains 3 blue and 7 green balls. If one ball is drawn from each bag, what is the Probability that two balls are of the same colour?
Answer: Solution: Let A be the event that a blue ball is drawn from each bag. Probability of drawing one blue ball out of 4 blue balls where there are a total of 9 balls in the first bag and that of drawing one blue ball out of 3 blue balls where there are a total of 10 balls in the second bag is \(P(A) = \frac{4}{9} \times \frac{3}{10}\) Let B be the event that a green ball is drawn from each bag. Probability of drawing one green ball out of 5 green balls where there are a total of 9 balls in the first bag and that of drawing one green ball out of 7 green balls where there are a total of 10 balls in the second bag is \(P(B) = \frac{5}{9} \times \frac{7}{10}\) Since both, the events are mutually exclusive and exhaustive events ∴ P(that both the balls are of the same colour) = P(both are of blue colour) or P(both are of green colour) = P(A) + P(B) \( = \frac{4}{9} \times \frac{3}{10} + \frac{5}{9} \times \frac{7}{10}\) \( = \frac{12}{90} + \frac{35}{90}\) \( = \frac{47}{90}\)
In simple words: This problem involves drawing one ball from each of two separate bags. To find the probability that both balls are the same color, we calculate the probability of both being blue and the probability of both being green, then add these probabilities since these are mutually exclusive events.

🎯 Exam Tip: When events occur from independent sources (like two different bags), their probabilities multiply. If multiple distinct scenarios lead to the desired outcome (e.g., both blue OR both green), sum their individual probabilities.

Question 13. Two cards are drawn one after the other from a pack of 52 cards with replacement. What is the probability that both the cards are drawn are face cards?
Answer: Solution: Two cards are drawn from a pack of 52 cards with replacement. ∴ n(S) = \(52 \times 52 = 2704\) Let A be the event that two cards drawn are face cards. First card from 12 face cards is drawn with replacement in \(^{12}C_1 = 12\) ways and second face card is drawn from 12 face card in \(^{12}C_1 = 12\) ways after replacement. ∴ n(A) = \(12 \times 12 = 144\) ∴ P(that both the cards drawn are face cards) = P(A) \( = \frac{n(A)}{n(S)} = \frac{12 \times 12}{52 \times 52} = \frac{144}{2704} = \frac{9}{169}\)
In simple words: This problem asks for the probability of drawing two face cards consecutively with replacement. Since the cards are replaced, each draw is an independent event, so we multiply the probability of drawing a face card in the first draw by the probability of drawing a face card in the second draw.

🎯 Exam Tip: For problems with replacement, probabilities for each draw remain constant because the total sample space is restored. There are 12 face cards (Jack, Queen, King of each of the 4 suits) in a standard 52-card deck.