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A problem is given to three students whose probabilities of solving it are \frac{1}{3},\frac{1}{4} and \frac{1}{6}  respectively. If the events of solving the problem are independent, find the probability that at least one of them solves it.

 

 

Answers (1)

E_{1}:$ Student 1 is able to solve the problem

P(E_{1}) = \frac{1}{3}

E_{2}:$ Student 2 is able to solve the problem

P(E_{2}) = \frac{1}{4}

E_{3}:$ Student 3 is able to solve the problem

P(E_{3}) = \frac{1}{6}

\\ P($Atlest one will able to solve $) =1- P(E'_{1}) \cdot P(E'_{2}) \cdot P(E'_{3})\\\\ = 1- \frac{2}{3} \times \frac{3}{4} \times \frac{5}{6} \\\\= \frac{72-30}{72} \\\\ = \frac{7}{12}

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Safeer PP

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