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Vince buys a box of candy that consists of six chocolate pieces, four fruit pieces and two mint pieces. He selects three pieces of candy at random without replacement.

  1. Calculate the probability that the first piece selected will be fruit flavored and the other two will be mint.

  2. Calculate the probability that all three pieces selected will be the same type of candy.

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Do you have any preliminary thoughts? Perhaps, some rough sketch or some preliminary thoughts. If you could elaborate a little on your preliminary thoughts you will get better help. – response May 22 '13 at 18:06

Since there are a total of $6+4+2=12$ candies and candies are not replaced, notice that: $$\begin{align*} P(\text{fruit} \to \text{mint} \to \text{mint}) &= P(\text{fruit})P(\text{mint | fruit})P(\text{mint | fruit,mint}) \\ &= \frac{4}{12}\cdot\frac{2}{11}\cdot\frac{1}{10}\\ &= \frac{1}{3}\cdot\frac{1}{11}\cdot\frac{1}{5}\\ &= \frac{1}{165} \end{align*}$$

Can you see why the $12$ decreased to $11$ and then to $10$? Can you see why the $2$ decreased to $1$?

Side Note: If replacement was allowed, then we would instead calculate: $$\frac{4}{12}\cdot\frac{2}{12}\cdot\frac{2}{12}$$

See if you can do the second one.

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Hint: For the first part, employ negative binomial distribution. For the second part, employ hypergeometric distribution.

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Hints for the first question

  1. In how many ways can Vince select 3 pieces of candy without replacement?

  2. In how many ways can Vince select 3 pieces of candy such that 1 is fruit flavored and the other two mint?

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Order matters in 2. – Maazul May 22 '13 at 18:10

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