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A shooter shoots at eight identical targets in a shooting gallery. The probability of hitting each target with each shot is the same. It takes the shooter 11 shots to hit all 8 targets. What is the probability that the shooter hit fewer than 4 targets with the first five shots?

I did the following: $p = \frac{8}{11}$, $q = \frac{3}{11}$ - for Bernoulli formula. Then I calculated $P(x<4) = 1 - P(x=4) - P(x=5) = 0.72$. But my friend told me that answer is $\frac{1}{2}$. Where is a mistake in my solution?

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  • $\begingroup$ You don't give sufficient detail in your answer. When you write $P(x=4)$ is that the probability of hitting four given 5 shots or 11 shots? $\endgroup$ Apr 23 at 22:39

2 Answers 2

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You don't know that the probability of hitting a target is in fact $\frac{8}{11}$. Indeed, you don’t have enough information to determine the probability of hitting a target. That's why your calculation is incorrect. Also, this is a conditional probability problem -- given that you needed $11$ shots to hit $8$ targets, what is the probability that the first $5$ shots hit fewer than $4$ targets?

But this problem should be solved via symmetry. Because it took $11$ shots to hit $8$ targets, you know that after $10$ shots the shooter hit exactly $7$ targets. By symmetry, the odds that more than half of those $7$ targets were hit in the first $5$ shots is exactly equal to the odds that more than half of those $7$ targets were hit in the second $5$ shots. Thus, the probability that the shooter hit $3$ or fewer targets in the first $5$ shots is exactly $\frac 12$.

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I consider two cases on the information we have and apply the counting method to find the probability considering that the outcomes of the experiment are equally likely (as the number of successes is fixed in both cases), which can be also used to solve the general case (the shooter hit fewer than $m$ targets with the first $n$ shots) where the symmetry argument does not work.

Case 1. We know that in 11 shots, the 8 targets are hit.

$$\frac{\binom{5}{2}\binom{6}{6}+\binom{5}{3}\binom{6}{5}}{\binom{11}{8}}=\frac{14}{33}.$$

Case 2. We know the 8th target is hit at the 11th shot.

$$\frac{\binom{5}{2}\binom{5}{5}+\binom{5}{3}\binom{5}{4}}{\binom{10}{7}}=\frac 12.$$

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