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(D) Let $p = \frac{1}{4}$ be the probability of hitting the target and $q = 1 - p = \frac{3}{4}$ be the probability of missing the target. For the sho

Vedclass · Accepted Answer

$(0.0658, 0.0660)$

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(D) Let $p = \frac{1}{4}$ be the probability of hitting the target and $q = 1 - p = \frac{3}{4}$ be the probability of missing the target. For the shooter to fire exactly $6$ bullets and hit the target exactly $3$ times,the $6^{th}$ bullet must be the $3^{rd}$ successful hit. This means in the first $5$ shots,the shooter must have hit the target exactly $2$ times and missed $3$ times. The probability of this event is given by the negative binomial distribution logic: $P = \binom{5}{2} p^2 q^3 	imes p = \binom{5}{2} p^3 q^3$. Substituting the values: $P = 10 	imes \left(\frac{1}{4}ight)^3 	imes \left(\frac{3}{4}ight)^3 = 10 	imes \frac{1}{64} 	imes \frac{27}{64} = \frac{270}{4096}$. Calculating the decimal value: $P = \frac{270}{4096} \approx 0.0659179$. This value lies in the interval $(0.0658, 0.0660)$.

$A$ 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

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