Conditional probability with pigs and cherries A bowl contains twenty cherries, exactly fifteen of which have had their stones REMOVED. A greedy pig eats five whole cherries without remarking on the presence or absence of stones. Subsequently, a cherry is picked randomly from the remaining fifteen.
(a) What is the probability that this cherry contains a stone?
(b) Given that this cherry contains a stone, what is the probability that the pig consumed at least one stone?
For (a), I thought I would condition on the pig eating 0, 1, ..., 5 stoned cherries, but I'm not sure if it is correct. As for (b), I'm not sure how to proceed.
 A: Here is an alternative solution method assuming that the pig eats first.
(a) Let $S=$ the number of cherries with $S$tones eaten by the pig. Then $P$($S$ = $i$) = $\dfrac{\binom{5}{i}\binom{15}{5-i}}{\binom{20}{5}}.$
Let $P$($S_6$ | $S = i$) be the probability of selecting a cherry with a $S$tone on the 6th draw given that the pig has eaten $i$ stones out of 5 cherries.
$P(S_6)=\sum_{i=0}^5$ $P$($S_6$ | $S = i$) $P$($S$ = $i$) = $\sum_{i=0}^5$ $\dfrac{\binom{5}{i}\binom{15}{5-i}}{\binom{20}{5}}$ $\dfrac{5-i}{15}=0.25$
(b) Correcting a typo in the answer of @Bananarama: $1-\dfrac{\binom{4}{0}\binom{15}{5}}{\binom{19}{5}}$ = 74.2%.
A: Suppose the cherry was picked before and removed and the pig ate after the cherry was removed, then it is clear:
(a) the probability the cherry contains a stone is $\frac{5}{20}=\frac{1}{4}$
(b) if the cherry contains a stone then there are now $19$ cherries in the bowl, $4$ of which contain stones, so the probability the pig eats at least $1$ cherry with stone is $1$ minus the probability the pig eats $5$ cherries without stones, there are 14 cherries with no stone, so that probability is $\dfrac{\binom{14}{5}}{\binom{19}{5}}$. So the probability we are looking for is $1-\dfrac{\binom{14}{5}}{\binom{19}{5}}\approx 83\%$
