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A shotgun has 6 slots and there is 1 bullet in it. The first person pulls the trigger and nothing was fired. The second person pulls the trigger without shuffling the chamber. What's probability that the second person dies?

Assume $P(A) = 1/6$ is the probability that the second person dies from firing the shotgun and $P(B)$ is the probability the first person doesn't die, so $P(B) = 5/6$. Then $P(B|A) = 1$ because the second person dies. Then

$P(A|B) = \frac{P(B|A) P(A)}{P(B)} = 1/5$

Is that right?

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Yes, that's right. You could also find the same result by counting the number of slots left that might contain the bullet, each of which has equal probability to contain the bullet.