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.
Calculate the probability that the first piece selected will be fruit flavored and the other two will be mint.
Calculate the probability that all three pieces selected will be the same type of candy.
Best Answer
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 $1$?
See if you can do the second one.