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

Std 11 Maths 2 Miscellaneous Exercise 7 Solutions Commerce Maths

Question 1. From a group of 2 men (M1, M2) and three women (W1, W2, W3), two persons are selected. Describe the sample space of the experiment. If E is the event in which one man and one woman are selected, then which are the cases favourable to E?
Answer: Solution: Let S be the sample space of the given event. \(\therefore\) S = {(M1, M2), (M1, W₁), (M1, W2), (M1, W3), (M2, W₁), (M2, W2), (M2, W3), (W1, W2), (W1, W3), (W2, W3)} Let E be the event that one man and one woman are selected. \(\therefore\) E = {(M1, W₁), (M1, W2), (M1, W3), (M2, W1), (M2, W2), (M2, W3)} Here, the order is not important in which 2 persons are selected e.g. (M1, M2) is the same as (M2, M1)
In simple words: The sample space (S) lists all possible pairs of people that can be selected from the given group. The event (E) specifically lists the pairs that consist of exactly one man and one woman.

🎯 Exam Tip: When forming a sample space for selections, ensure all unique combinations are listed. For events, clearly identify the specific outcomes that satisfy the given condition.

 

Question 2. Three groups of children contain respectively 3 girls and 1 boy, 2 girls and 2 boys and 1 girl and 3 boys. One child is selected at random from each group. What is the chance that the three selected consist of 1 girl and 2 boys?
Answer: Solution:

Group - I	Group - II	Group - III
Girls	Boys	Girls	Boys	Girls	Boys
3	1	2	2	1	3

Let G1, G2, G3 denote events for selecting a girl, and B1, B2, B3 denote events for selecting a boy from 1st, 2nd and 3rd groups respectively. Then P(G1) = \(\frac{3}{4}\), P(G2) = \(\frac{2}{4}\), P(G3) = \(\frac{1}{4}\) P(B1) = \(\frac{1}{4}\), P(B2) = \(\frac{2}{4}\), P(B3) = \(\frac{3}{4}\) Where G1, G2, G3, B1, B2 and B3 are mutually exclusive events. Let E be the event that 1 girl and 2 boys are selected \(\therefore\) E = (G1 \(\cap\) B2 \(\cap\) B3) U (B₁ \(\cap\) G2 \(\cap\) B3) U (B1 \(\cap\) B2 \(\cap\) G3) \(\therefore\) P(E) = P(G₁ \(\cap\) B2 \(\cap\) B3) + P(B₁ \(\cap\) G2 \(\cap\) B3) + P(B1 \(\cap\) B2 \(\cap\) G3) \(=\left(\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4}\right)+\left(\frac{1}{4} \times \frac{2}{4} \times \frac{3}{4}\right)+\left(\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4}\right)\) \(=\frac{18+6+2}{64}\) \(=\frac{26}{64}\) \(=\frac{13}{32}\)
In simple words: We calculate the probability of selecting one girl and two boys by considering all possible combinations of a girl from one group and boys from the other two, then summing these probabilities.

🎯 Exam Tip: Remember to consider all distinct cases that satisfy the event condition (e.g., girl from group 1, boys from groups 2 & 3; or boy from group 1, girl from group 2, boy from group 3, etc.) when events are mutually exclusive.

 

Question 3. A room has 3 sockets for lamps. From a collection of 10 light bulbs, 6 are defective. A person selects 3 at random and puts them in every socket. What is the probability that the room, will be lit?
Answer: Solution: Total number of bulbs = 10 Number of defective bulbs = 6 \(\therefore\) Number of non-defective bulbs = 4 3 bulbs can be selected out of 10 light bulbs in \(^{10}C_3\) ways. \(\therefore\) n(S) = \(^{10}C_3\) Let A be the event that room is lit. \(\therefore\) A' is the event that the room is not lit. For A' the bulbs should be selected from the 6 defective bulbs. This can be done in \(^6C_3\) ways. \(\therefore\) n(A') = \(^6C_3\) \(\therefore\) P(A') = \(\frac{n(A')}{n(S)} = \frac{^6C_3}{^{10}C_3}\) \(\therefore\) P(Room is lit) = 1 – P(Room is not lit) \(\therefore\) P(A) = 1 - P(A')
\(=\) \(1-\frac{^6C_3}{^{10}C_3}\)
\(=\) \(1-\frac{6\times5\times4}{10\times9\times8}\)
\(=\) \(1-\frac{1}{6}\)
\(=\frac{5}{6}\)
In simple words: To find the probability that the room is lit, it's easier to find the probability that the room is NOT lit (i.e., all 3 selected bulbs are defective) and subtract that from 1.

🎯 Exam Tip: For "at least one" type probability problems, it's often simpler to calculate the probability of the complementary event ("none") and subtract it from 1. This avoids complex case enumeration.

 

Question 4. There are 2 red and 3 black balls in a bag. 3 balls are taken out at random from the bag. Find the probability of getting 2 red and 1 black ball or 1 red and 2 black balls.
Answer: Solution: There are 2 + 3 = 5 balls in the bag and 3 balls can be drawn out of these in \(^5C_3 = \frac{5\times4\times3}{1\times2\times3} = 10\) ways. \(\therefore\) n(S) = 10 Let A be the event that 2 balls are red and 1 ball is black 2 red balls can be drawn out of 2 red balls in \(^2C_2 = 1\) way and 1 black ball can be drawn out of 3 black balls in \(^3C_1 = 3\) ways. \(\therefore\) n(A) = \(^2C_2 \times ^3C_1 = 1 \times 3 = 3\) \(\therefore\) P(A) = \(\frac{n(A)}{n(S)} = \frac{3}{10}\) Let B be the event that 1 ball is red and 2 balls are black 1 red ball out of 2 red balls can be drawn in \(^2C_1 = 2\) ways and 2 black balls out of 3 black balls can be drawn in \(^3C_2 = \frac{3\times2}{1\times2} = 3\) ways. \(\therefore\) n(B) = \(^2C_1 \times ^3C_2 = 2 \times 3 = 6\) \(\therefore\) P(B) = \(\frac{n(B)}{n(S)} = \frac{6}{10}\) Since A and B are mutually exclusive and exhaustive events \(\therefore\) P(A \(\cap\) B) = 0 \(\therefore\) Required probability = P(A \(\cup\) B) = P(A) + P(B)
\(=\) \(\frac{3}{10}+\frac{6}{10}\)
\(=\) \(\frac{9}{10}\)
In simple words: We calculate the probability of two separate events (2 red, 1 black OR 1 red, 2 black) and since these events cannot happen at the same time (mutually exclusive), we simply add their individual probabilities.

🎯 Exam Tip: For "OR" conditions in probability problems involving mutually exclusive events, always remember to sum their individual probabilities. Ensure you use combinations (\(^{n}C_r\)) when the order of selection doesn't matter.

 

Question 5. A box contains 25 tickets numbered 1 to 25. Two tickets are drawn at random. What is the probability that the product of the numbers is even?
Answer: Solution: Two tickets can be drawn out of 25 tickets in \(^{25}C_2 = \frac{25\times24}{1\times2} = 300\) ways. \(\therefore\) n(S) = 300 Let A be the event that product of two numbers is even. This is possible if both numbers are even, or one number is even and other is odd. As there are 13 odd numbers and 12 even numbers from 1 to 25. \(\therefore\) n(A) = \(^{12}C_2 + ^{12}C_1 \times ^{13}C_1\)
\(=\) \(\frac{12\times11}{1\times2} + 12 \times 13\)
\(=\) \(66 + 156\)
\(=\) \(222\) \(\therefore\) Required probability = P(A)
\(=\) \(\frac{n(A)}{n(S)}\)
\(=\) \(\frac{222}{300}\)
\(=\) \(\frac{37}{50}\)
In simple words: The product of two numbers is even if at least one of the numbers is even. We calculate the number of ways to pick two even numbers, plus the number of ways to pick one even and one odd number, then divide by the total ways to pick two numbers.

🎯 Exam Tip: Remember the rules for products of even/odd numbers: Even \(\times\) Even = Even, Even \(\times\) Odd = Even, Odd \(\times\) Odd = Odd. This helps categorize favorable outcomes correctly.

 

Question 6. A, B and C are mutually exclusive and exhaustive events associated with the random experiment. Find P(A), given that P(B) = \(\frac{3}{2}\) P(A) and P(C) = \(\frac{1}{2}\) P(B)
Answer: Solution: P(B) = \(\frac{3}{2}\) P(A) and P(C) = \(\frac{1}{2}\) P(B) Since A, B, C are mutually exclusive and exhaustive events, \(\therefore\) P(A \(\cup\) B \(\cup\) C) = P(A) + P(B) + P(C) = 1 \(\therefore\) P(A) + \(\frac{3}{2}\) P(A) + P(B) = 1 \(\therefore\) P(A) + \(\frac{3}{2}\) P(A) + \(\frac{1}{2}\) \(\times\) \(\frac{3}{2}\) P(A) = 1 \(\therefore\) P(A) + \(\frac{3}{2}\) P(A) + \(\frac{3}{4}\) P(A) = 1 \(\therefore\) P(A) \(\times\) \(\left(1+\frac{3}{2}+\frac{3}{4}\right) = 1\) \(\therefore\) P(A) \(\times\) \(\left(\frac{13}{4}\right) = 1\) \(\therefore\) P(A) = \(\frac{4}{13}\)
In simple words: Since A, B, and C cover all possibilities and don't overlap, their probabilities must sum to 1. We use the given relationships between P(A), P(B), and P(C) to express everything in terms of P(A) and then solve for it.

🎯 Exam Tip: When events are mutually exclusive and exhaustive, their probabilities add up to 1. Substitute the given relationships to form an equation with a single unknown probability.

 

Question 7. An urn contains four tickets marked with numbers 112, 121, 122, 222, and one ticket is drawn at random. Let Aᵢ (i = 1, 2, 3) be the event that ith digit of the number of the ticket drawn is 1. Discuss the independence of the events A1, A2, and A3.
Answer: Solution: One ticket can be drawn out of 4 tickets in \(^4C_1 = 4\) ways. \(\therefore\) n(S) = 4 According to the given information, Let A₁ be the event that 1st digit of the number of tickets is 1 A2 be the event that the 2nd digit of the number of tickets is 1 A3 be the event that the 3rd digit of the number of tickets is 1 \(\therefore\) A₁ = {112, 121, 122}, A2 = {112}, A3 = {121} \(\therefore\) P(A₁) = \(\frac{n(A_1)}{n(S)} = \frac{3}{4}\) P(A2) = \(\frac{n(A_2)}{n(S)} = \frac{1}{4}\) P(A3) = \(\frac{n(A_3)}{n(S)} = \frac{1}{4}\) P(A₁) P(A2) = \(\frac{3}{16}\) P(A2) P(A3) = \(\frac{1}{16}\) ...(i) P(A₁) P(A3) = \(\frac{3}{16}\) A₁ \(\cap\) A2 = {112}, A2 \(\cap\) A3 = \(\phi\), A1 \(\cap\) A3 = {121} P(A₁ \(\cap\) A2) = \(\frac{n(A_1 \cap A_2)}{n(S)} = \frac{1}{4}\) P(A2 \(\cap\) A3) = 0, ...(ii) P(A₁ \(\cap\) A3) = \(\frac{1}{4}\) \(\therefore\) From (i) and (ii), P(A₁).P(A2) \(\neq\) P(A₁ \(\cap\) A2), P(A2).P(A3) \(\neq\) P(A2 \(\cap\) A3), ....(iii) P(A₁).P(A3) \(\neq\) P(A₁ \(\cap\) A3) \(\therefore\) A1, A2, A3 are not pairwise independent For mutual independence of events A1, A2, A3 We require to have P(A1 \(\cap\) A2 \(\cap\) A3) = P(A1) P(A2) P(A3) and P(A1) P(A2) = P(A1 \(\cap\) A2), P(A2) P(A3) = P(A2 \(\cap\) A3), P(A1) P(A3) = P(A1 \(\cap\) A3) \(\therefore\) From (iii), A1, A2, A3 are not mutually independent.
In simple words: To check for independence, we compare the product of individual probabilities with the probability of their intersection. If these are not equal for any pair of events, then the events are not pairwise independent, and therefore not mutually independent.

🎯 Exam Tip: For events to be independent, P(A \(\cap\) B) must equal P(A)P(B). For three events, you must check all pairwise intersections and the intersection of all three for mutual independence.

 

Question 8. The odds against a certain event are 5 : 2 and the odds in favour of another independent event are 6 : 5. Find the chance that at least one of the events will happen.
Answer: Solution: Let A and B be two independent events. Odds against A are 5 : 2 \(\therefore\) the probability of occurrence of event A is given by P(A) = \(\frac{2}{5+2} = \frac{2}{7}\) Odds in favour of B are 6:5 \(\therefore\) the probability of occurrence of event B is given by P(B) = \(\frac{6}{6+5} = \frac{6}{11}\) \(\therefore\) P(at least one event will happen) = P(A \(\cup\) B)
\(=\) P(A) + P(B) – P(A \(\cap\) B)
\(=\) P(A) + P(B) – P(A) P(B) ......[. A and B are independent events]
\(=\) \(\frac{2}{7}+\frac{6}{11}-\frac{2}{7}\times\frac{6}{11}\)
\(=\) \(\frac{2}{7}+\frac{6}{11}-\frac{12}{77}\)
\(=\) \(\frac{22+42-12}{77}\)
\(=\) \(\frac{52}{77}\)
In simple words: First, convert the given odds into probabilities for each event. Then, since the events are independent, the probability that at least one happens is found by P(A) + P(B) - P(A)P(B).

🎯 Exam Tip: Remember that "odds against a : b" means probability is b/(a+b), and "odds in favour of a : b" means probability is a/(a+b). For independent events, P(A \(\cap\) B) = P(A)P(B).

 

Question 9. The odds against a husband who is 55 years old living till he is 75 is 8 : 5 and it is 4 : 3 against his wife who is now 48, living till she is 68. Find the probability that (i) the couple will be alive 20 years hence (ii) at least one of them will be alive 20 years hence.
Answer: Solution: Let A be the event that husband would be alive after 20 years. Odds against A are 8:5 \(\therefore\) the probability of occurrence of event A is given by P(A) = \(\frac{5}{8+5} = \frac{5}{13}\) \(\therefore\) P(A') = 1 – P(A)
\(=\) \(1-\frac{5}{13}\)
\(=\) \(\frac{8}{13}\) Let B be the event that wife would be alive after 20 years. Odds against B are 4 : 3 \(\therefore\) the probability of occurrence of event B is given by P(B) = \(\frac{3}{4+3} = \frac{3}{7}\) \(\therefore\) P(B') = 1 – P(B)
\(=\) \(1-\frac{3}{7}\)
\(=\) \(\frac{4}{7}\) Since A and B are independent events \(\therefore\) A' and B' are also independent events (i) Let X be the event that both will be alive after 20 years. \(\therefore\) P(X) = (A \(\cap\) B) \(\therefore\) P(X) = P(A) . P(B)
\(=\) \(\frac{5}{13}\times\frac{3}{7}\)
\(=\) \(\frac{15}{91}\) (ii) Let Y be the event that at least one will be alive after 20 years. \(\therefore\) P(Y) = P(at least one would be alive)
\(=\) \(1 – P(both would not be alive)\)
\(=\) \(1 – P(A' \cap Β')\)
\(=\) \(1 – P(A'). P(Β')\)
\(=\) \(1 – \frac{8}{13}\times\frac{4}{7}\)
\(=\) \(1 – \frac{32}{91}\)
\(=\) \(\frac{59}{91}\)
In simple words: First, convert the odds for the husband and wife surviving into probabilities. For both to be alive, multiply their individual survival probabilities. For at least one to be alive, subtract the probability that neither is alive from 1.

🎯 Exam Tip: Always convert odds into probabilities first. For independent events: P(A and B) = P(A)P(B); P(A or B) = 1 - P(A' and B').

 

Question 10. Two throws are made, the first with 3 dice and the second with 2 dice. The faces of each die are marked with the number 1 to 6. What is the probability that the total in the first throw is not less than 15 and at the same time the total in the second throw is not less than 8?
Answer: Solution: When 3 dice are thrown, then the sample space S₁ has 6 \(\times\) 6 \(\times\) 6 = 216 sample points. \(\therefore\) n(S₁) = 216 Let A be the event that the sum of the numbers is not less than 15. \(\therefore\) A = {(3, 6, 6), (4, 5, 6), (4, 6, 5), (4, 6, 6), (5, 4, 6), (5, 5, 5), (5, 5, 6), (5, 6, 4), (5, 6, 5), (5, 6, 6), (6, 3, 6), (6, 4, 5), (6, 4, 6), (6, 5, 4), (6, 5, 5), (6, 5, 6), (6, 6, 3), (6, 6, 4), (6, 6, 5), (6, 6, 6)} \(\therefore\) n(A) = 20 \(\therefore\) P(A) = \(\frac{n(A)}{n(S_1)} = \frac{20}{216} = \frac{5}{54}\) When 2 dice are thrown, the sample space S2 has 6 \(\times\) 6 = 36 sample points. \(\therefore\) n(S2) = 36 Let B be the event that sum of numbers is not less than 8. \(\therefore\) B = {(2, 6), (3, 5), (3,6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} \(\therefore\) n(B) = 15 \(\therefore\) P(B) = \(\frac{n(B)}{n(S_2)} = \frac{15}{36} = \frac{5}{12}\) A \(\cap\) B = Event that the total in the first throw is not less than 15 and at the same time the total in the second throw is not less than 8. \(\therefore\) A and B are independent events \(\therefore\) P(A \(\cap\) B) = P(A) . P(B)
\(=\) \(\frac{5}{54}\times\frac{5}{12}\)
\(=\) \(\frac{25}{648}\)
In simple words: Calculate the probability for each throw separately: the sum not less than 15 for three dice (event A) and the sum not less than 8 for two dice (event B). Since the throws are independent, multiply these two probabilities to find the combined chance.

🎯 Exam Tip: For problems involving multiple independent events (like separate dice throws), calculate the probability of each event individually and then multiply them to find the probability of both occurring.

 

Question 11. Two-thirds of the students in a class are boys and the rest are girls. It is known that the probability of a girl getting first class is 0.25 and that of a boy getting is 0.28. Find the probability that a student chosen at random will get first class.
Answer: Solution: Let A be the event that student chosen is a boy B be the event that student chosen is a girl C be the event that student gets first class \(\therefore\) P(A) = \(\frac{2}{3}\), P(B) = \(\frac{1}{3}\) Probability of student getting first class, given that student is boy Probability of student getting first class given that student is a girl, is P(C/A) = 0.28 = \(\frac{28}{100}\) and P(C/B) = 0.25 = \(\frac{25}{100}\) \(\therefore\) Required probability = P((A \(\cap\) C) \(\cup\) (B \(\cap\) C)) Since A \(\cap\) C and B \(\cap\) C are mutually exclusive events \(\therefore\) Required probability = P(A \(\cap\) C) + P(B \(\cap\) C)
\(=\) P(A) . P(C/A) + P(B) . P(C/B)
\(=\) \(\frac{2}{3} \times \frac{28}{100} + \frac{1}{3} \times \frac{25}{100}\)
\(=\) \(\frac{56+25}{300}\)
\(=\) \(\frac{81}{300}\)
\(=\) \(0.27\)
In simple words: We calculate the probability of a boy getting first class (P(Boy) \(\times\) P(First Class | Boy)) and the probability of a girl getting first class (P(Girl) \(\times\) P(First Class | Girl)). Since these are mutually exclusive, we add them to get the total probability of a randomly chosen student getting first class.

🎯 Exam Tip: This problem uses the Law of Total Probability. Remember to correctly identify the prior probabilities (P(Boy), P(Girl)) and the conditional probabilities (P(First Class | Boy), P(First Class | Girl)).

 

Question 12. A number of two digits is formed using the digits 1, 2, 3,......, 9. What is the probability that the number so chosen is even and less than 60?
Answer: Solution: The number of two digits can be formed from the given 9 digits in 9 \(\times\) 9 = 81 different ways. \(\therefore\) n(S) = 81 Let A be the event that the number is even and less than 60. Since the number is even, the unit place of two digits can be filled in \(^4P_1 = 4\) different ways by any one of the digits 2, 4, 6, 8. Also the number is less than 60, so tenth place can be filled in \(^5P_1 = 5\) different ways by any one of the digits 1, 2, 3, 4, 5. \(\therefore\) n(A) = 4 \(\times\) 5 = 20 \(\therefore\) Required probability = P(A) = \(\frac{n(A)}{n(S)} = \frac{20}{81}\)
In simple words: First, find the total number of two-digit numbers possible. Then, count the numbers that are both even (unit digit is 2, 4, 6, or 8) and less than 60 (tens digit is 1, 2, 3, 4, or 5). The probability is the ratio of favorable outcomes to total outcomes.

🎯 Exam Tip: Carefully list the constraints for each digit when counting favorable outcomes. The "and" condition (even AND less than 60) means both conditions must be met simultaneously for the number.

 

Question 13. A bag contains 8 red balls and 5 white balls. Two successive draws of 3 balls each are made without replacement. Find the probability that the first drawing will give 3 white balls and the second drawing will give 3 red balls.
Answer: Solution: Total number of balls = 8 + 5 = 13. 3 balls can be drawn out of 13 balls in \(^{13}C_3\) ways. \(\therefore\) n(S) = \(^{13}C_3\) Let A be the event that all 3 balls drawn are white. 3 white balls can be drawn out of 5 white balls in \(^5C_3\) ways. \(\therefore\) n(A) = \(^5C_3\) \(\therefore\) P(A) = \(\frac{n(A)}{n(S)} = \frac{^5C_3}{^{13}C_3} = \frac{5\times4\times3}{13\times12\times11} = \frac{5}{143}\) After drawing 3 white balls which are not replaced in the bag, there are 10 balls left in the bag out of which 8 are red balls. Let B be the event that the second draw of 3 balls are red. \(\therefore\) Probability of drawing 3 red balls, given that 3 white balls have been already drawn, is given by P(B/A) = \(\frac{^8C_3}{^{10}C_3} = \frac{8\times7\times6}{10\times9\times8} = \frac{7}{15}\) \(\therefore\) Required probability = P(A \(\cap\) B)
\(=\) P(A) . P(B/A)
\(=\) \(\frac{5}{143}\times\frac{7}{15}\)
\(=\) \(\frac{7}{429}\)
In simple words: This is a conditional probability problem without replacement. First, calculate the probability of drawing 3 white balls. Then, adjust the total number of balls and red balls remaining, and calculate the probability of drawing 3 red balls from the remaining. Multiply these two probabilities together.

🎯 Exam Tip: For "without replacement" problems, remember to update the total number of items and the number of favorable items for successive draws before calculating the conditional probability.

 

Question 14. The odds against student X solving a business statistics problem are 8 : 6 and the odds in favour of student Y solving the same problem are 14 : 16 (i) What is the chance that the problem will be solved, if they try independently? (ii) What is the probability that neither solves the problem?
Answer: Solution: (i) Let A be the event that X solves the problem B be the event that Y solves the problem. Since the odds against student X solving the problem are 8:6 \(\therefore\) Probability of occurrence of event A is given by P(A) = \(\frac{6}{8+6} = \frac{6}{14}\) and P(A') = 1 – P(A)
\(=\) \(1-\frac{6}{14}\)
\(=\) \(\frac{8}{14}\) Also, the odds in favour of student Y solving the problem are 14 : 16 \(\therefore\) Probability of occurrence of event B is given by P(B) = \(\frac{14}{14+16} = \frac{14}{30}\) and P(B') = 1 – P(B)
\(=\) \(1-\frac{14}{30}\)
\(=\) \(\frac{16}{30}\) Now A and B are independent events. \(\therefore\) A' and B' are also independent events \(\therefore\) A' \(\cap\) B' = Event that neither solves the problem
\(=\) P(A' \cap Β')
\(=\) P(A') . P(Β')
\(=\) \(\frac{8}{14}\times\frac{16}{30}\)
\(=\) \(\frac{32}{105}\) A \(\cup\) B = the event that the problem is solved \(\therefore\) P(problem will be solved) = P(A \(\cup\) B)
\(=\) \(1 – P(A \cup Β)'\)
\(=\) \(1 – P(A' \cap Β')\)
\(=\) \(1-\frac{32}{105}\)
\(=\) \(\frac{73}{105}\) (ii) P (neither solves the problem) = P(A' \(\cap\) Β')
\(=\) P(A') P(Β')
\(=\) \(\frac{8}{14}\times\frac{16}{30}\)
\(=\) \(\frac{32}{105}\)
In simple words: First, convert odds to probabilities for X solving (P(A)) and Y solving (P(B)). (i) For the problem to be solved (at least one solves it), calculate 1 minus the probability that *neither* solves it. (ii) For neither to solve it, find the probability that X doesn't solve it (P(A')) and Y doesn't solve it (P(B')), then multiply these since they are independent.

🎯 Exam Tip: Carefully convert "odds against" and "odds in favour" to actual probabilities. When dealing with independent events and "at least one" scenarios, the complementary event approach (1 - P(none)) is often the most efficient.