Test Index

Probability

Question : 102 of 127

Marks: +1, -0

C5

Solution:

Concept:

Find the probability that exactly two of the three shooters hit the target.

Explanation:

Given probabilities:

P(A) = \frac{3}{4}

P(B) = \frac{4}{5}

P(C) = \frac{5}{6}

Their miss probabilities:

P(\bar{A}) = \frac{1}{4}

P(\bar{B}) = \frac{1}{5}

P(\bar{C}) = \frac{1}{6}

Exactly two hits occur in three mutually exclusive cases:

1. A and B hit, C misses:

\frac{3}{4} \times \frac{4}{5} \times \frac{1}{6} = \frac{12}{120} = \frac{1}{10}

2. A and C hit, B misses:

\frac{3}{4} \times \frac{1}{5} \times \frac{5}{6} = \frac{15}{120} = \frac{1}{8}

3. B and C hit, A misses:

\frac{1}{4} \times \frac{4}{5} \times \frac{5}{6} = \frac{20}{120} = \frac{1}{6}

Sum:

\frac{1}{10} + \frac{1}{8} + \frac{1}{6} = \frac{12 + 15 + 20}{120} = \frac{47}{120}

Answer:

\frac{47}{120}

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