Question

The probability of solving a problem by three persons A, B and C independently is \(\frac{1}{2}\), \(\frac{1}{4}\) and \(\frac{1}{3}\) respectively. Then the probability of the problem is solved by any two of them is

• A. \(\frac{1}{12}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{1}{24}\)
• D. \(\frac{1}{8}\)

Answer 1

The Correct Option is B

Let $ P(A) = \frac{1}{2} $, $ P(B) = \frac{1}{4} $, and $ P(C) = \frac{1}{3} $. We want to find the probability that exactly two of them solve the problem. This can happen in three ways: $ A $ and $ B $ solve, $ A $ and $ C $ solve, or $ B $ and $ C $ solve.

$$ P(\text{exactly two}) = P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C). $$

Using independence, we calculate each term:

• $ P(C') = 1 - P(C) = 1 - \frac{1}{3} = \frac{2}{3} $,
• $ P(B') = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4} $,
• $ P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} $.

Now compute each term:

$$ P(A \cap B \cap C') = P(A) P(B) P(C') = \left(\frac{1}{2}\right)\left(\frac{1}{4}\right)\left(\frac{2}{3}\right) = \frac{2}{24}. $$ $$ P(A \cap B' \cap C) = P(A) P(B') P(C) = \left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\left(\frac{1}{3}\right) = \frac{3}{24}. $$ $$ P(A' \cap B \cap C) = P(A') P(B) P(C) = \left(\frac{1}{2}\right)\left(\frac{1}{4}\right)\left(\frac{1}{3}\right) = \frac{1}{24}. $$

Add these probabilities:

$$ P(\text{exactly two}) = \frac{2}{24} + \frac{3}{24} + \frac{1}{24} = \frac{6}{24} = \frac{1}{4}. $$

Answer 2

We want to find the probability that the problem is solved by any two of them. This can happen in three ways:

1. A and B solve it, but C doesn't: \(P(A \cap B \cap C') = P(A)P(B)P(C') = (\frac{1}{2})(\frac{1}{4})(1 - \frac{1}{3}) = (\frac{1}{8})(\frac{2}{3}) = \frac{1}{12}\)
2. A and C solve it, but B doesn't: \(P(A \cap B' \cap C) = P(A)P(B')P(C) = (\frac{1}{2})(1 - \frac{1}{4})(\frac{1}{3}) = (\frac{1}{2})(\frac{3}{4})(\frac{1}{3}) = \frac{1}{8}\)
3. B and C solve it, but A doesn't: \(P(A' \cap B \cap C) = P(A')P(B)P(C) = (1 - \frac{1}{2})(\frac{1}{4})(\frac{1}{3}) = (\frac{1}{2})(\frac{1}{4})(\frac{1}{3}) = \frac{1}{24}\)

Since these are mutually exclusive events, we can add their probabilities:

\(P(\text{any two solve it}) = P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C)\)

\(= \frac{1}{12} + \frac{1}{8} + \frac{1}{24}\)

\(= \frac{2 + 3 + 1}{24}\)

\(= \frac{6}{24}\)

\(= \frac{1}{4}\)