There is a bag of 8 candies, and 3 are chocolates. You eat candy until the chocolates are gone. What is the probability you will have eaten 7 candies? You buy a bag of $8$ candies, of which $3$ are chocolates, but all candies look alike. You eat candies from the bag until you have eaten all three chocolates. What is the probability you will have eaten exactly $7$ of the candies in the bag?
 A: A particular sequence of picks is just as likely as the same sequence in reverse. But now the question becomes:
What is the probability that the first pick is not a chocolate, and the second pick is?
So the answer is obviously $\dfrac{5}{8}\times\dfrac{3}{7}$.
A: As D.R. pointed out, there must be $2$ chocolates among the first $6$ candies. Choose these $2$ positions in $\binom{6}{2}$ ways. The $7^{th}$ position must be a chocolate, so in total, the positions of the chocolates can be chosen in $\binom{6}{2}$ ways. Hence, your required probability is $\binom{6}{2}/\binom{8}{3} = 0.2679$.
A: As others have pointed out, we must select $2$ chocolates and $4$ non-chocolates in the first $6$ selections. The probability that this occurs is 
$$\frac{{3\choose2} {5\choose4}}{8\choose6}$$
This comes from the hypergeometric distribution. Then there is one chocolate and one non-chocolate remaining so we then select the third chocolate with probability $\frac{1}{2}$. Hence the desired probability is
$$\frac{{3\choose2} {5\choose4}}{8\choose6}\cdot\frac{1}{2}\approx0.268$$ 
A: HINT: if the person stops after the 7th candy, then that one must have been the third chocolate. The other two can be anywhere in the first 6.
If the question is asking “at least 7 candies”, consider the case where the 8th candy is the third chocolate, and sum the two answers together.
