The probabilities of solving a problem by ‘3’ students A,B,C independently are 13,14,15respectively then the probability for

The problem will be solved by exactly one student is 13 30 , The problem will be solved by A,B but not C is 1 5

The probabilities of solving a problem by ‘3’ students A,B,C independently are 13,14,15respectively then the probability for

P ( A ) = 1 3 ; P ( B ) = 1 4 , P ( 1 5 ) ( A ) P ( A ∪ B ∪ C ) = 1 − P ( A ¯ ) . P ( B ¯ ) . P ( C ¯ ) = 1 − 2 3 . 3 4 . 4 5 = 3 5 ( B ) P ( A ∩ B ¯ ∩ C ¯ ) + P ( A ¯ ∩ B ∩ C ¯ ) + P ( A ¯ ∩ B ¯ ∩ C ) = 1 3 . 3 4 . 4 5 + 2 3 . 1 4 . 4 5 + 2 3 . 3 4 . 1 5 ( C ) P ( A ∩ B ∩ C ¯ ) = 1 3 . 1 4 . 4 5 = 1 15 ( D ) P ( A ¯ ∩ B ¯ ∩ C ¯ ) = 2 5

The problem will be solved by A,B but not C is 1 5

The problem will be solved is 3 5

The problem will not solved by any student is 3 5

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The probabilities of solving a problem by ‘3’ students A,B,C independently are $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$ respectively then the probability for

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The problem will be solved by exactly one student is $\frac{13}{30}$

The problem will be solved by A,B but not C is $\frac{1}{5}$

The problem will be solved is $\frac{3}{5}$

The problem will not solved by any student is $\frac{3}{5}$

answer is A, B.

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Detailed Solution

$\begin{array}{l} P (A) = \frac{1}{3}; P (B) = \frac{1}{4}, P (\frac{1}{5}) \\ (A) P (A \cup B \cup C) = 1 - P (\bar{A}) . P (\bar{B}) . P (\bar{C}) \\ = 1 - \frac{2}{3} . \frac{3}{4} . \frac{4}{5} = \frac{3}{5} \\ (B) P (A \cap \bar{B} \cap \bar{C}) + P (\bar{A} \cap B \cap \bar{C}) + P (\bar{A} \cap \bar{B} \cap C) \\ = \frac{1}{3} . \frac{3}{4} . \frac{4}{5} + \frac{2}{3} . \frac{1}{4} . \frac{4}{5} + \frac{2}{3} . \frac{3}{4} . \frac{1}{5} \end{array}$

$\begin{array}{l} (C) P (A \cap B \cap \bar{C}) = \frac{1}{3} . \frac{1}{4} . \frac{4}{5} = \frac{1}{15} \\ (D) P (\bar{A} \cap \bar{B} \cap \bar{C}) = \frac{2}{5} \end{array}$