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1. The probability of success of three students X,Y and Z in the one examination are $\frac{1}{5}$ , $\frac{1}{4}$ and $\frac{1}{3}$ respectively. Find the probability of success of at least two.

A. $\frac{1}{6}$

B. $\frac{2}{5}$

C. $\frac{3}{4}$

D. $\frac{3}{5}$

Solution:

P (X) = \frac{1}{5},

P (Y) = \frac{1}{4},

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

Required probability:

= [P (A) P (B) {1 - P (C)}]

+

[{1 - P (A)} P (B) P (C)] +

[P (A) P (C) {1 - P (B)}] +

[P (A) P (B) P (C)]

= [\frac{1}{4} \times \frac{1}{3} \times \frac{4}{5}] +

[\frac{3}{4} \times \frac{1}{3} \times \frac{1}{5}] +

[\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}] +

[\frac{1}{4} \times \frac{1}{3} \times \frac{1}{5}]

\begin{aligned} = \frac{4}{60} + \frac{3}{60} + \frac{2}{60} + \frac{1}{60} \\ = \frac{10}{60} \\ = \frac{1}{6} \end{aligned}

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