A soldier fires three bullets on enemy: The probability that the enemy will be killed by

\frac{1}{3}

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\frac{2}{3}

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\frac{7}{8}

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\frac{1}{8}

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Solution

The correct option is C $\frac{7}{8}$
The probability that the enemy will be killed =1- the probability of missing all three hits

$= 1 - {(\frac{1}{2})}^{3} = \frac{7}{8}$

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A soldier fires three bullets on enemy: The probability that the enemy will be killed by

The correct option is C 78The probability that the enemy will be killed =1- the probability of missing all three hits=1−(12)3=78

The correct option is C $\frac{7}{8}$
The probability that the enemy will be killed =1- the probability of missing all three hits

$= 1 - {(\frac{1}{2})}^{3} = \frac{7}{8}$