A shooter can hit a given target with probability $\frac{1}{4}$ . She keeps firing a bullet at the target until she hits it successfully three times and then she stops firing. The probability that she fires exactly six bullets lies in the interval.

(0.5272, 0.5274)

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(0.2636, 0.2638)

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(0.1317, 0.1319)

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(0.0658, 0.0660)

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Solution

The correct option is B $(0.0658, 0.0660)$
if she fires exactly six bullets, then the sixth bullet will be the third bullet which will hit the target,
so for that, we have to find the probability that in previous five bullets only two hit the target,

$P = \frac{5!}{3! 2!} (\frac{1}{4})^{2} (\frac{3}{4})^{3}$

$P = 0.065917$

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A shooter can hit a given target with probability 14. She keeps firing a bullet at the target until she hits it successfully three times and then she stops firing. The probability that she fires exactly six bullets lies in the interval.

The correct option is B (0.0658,0.0660)if she fires exactly six bullets, then the sixth bullet will be the third bullet which will hit the target, so for that, we have to find the probability that in previous five bullets only two hit the target,P=5!3!2!(14)2(34)3P=0.065917

A shooter can hit a given target with probability $\frac{1}{4}$ . She keeps firing a bullet at the target until she hits it successfully three times and then she stops firing. The probability that she fires exactly six bullets lies in the interval.