A package contains 16 candy canes, 5 of which are cracked. If 9 are selected, find the probability of getting no cracked candy canes.

Question

kirbykirby · Answer

You could think of this in two ways:

Method 1:
If 5 of the 16 candy canes are cracked, then 11 are not cracked. Thus:

$$ \begin{align} P(\text{9 not cracked})&=P(\text{1st not cracked})\cdot P(\text{2nd not cracked})\cdots P(\text{9th not cracked})\
&=\frac{11}{16}\times\frac{10}{15}\times \frac{9}{14}\times\cdots\times\frac{3}{8}\end{align}$$

Method 2:
Your sample space has $\large {16 \choose 9} $ outcomes since you choose 9 candy canes from the 16. Now, we're interested in only the non-cracked ones (11 of them). So, in the numerator, you'd have $\large {11\choose 9} $ is the number of outcomes of the event "getting none cracked" since you choose 9 out of 11 of the non cracked ones. 

$$ P(\text{9 not cracked})=\large \frac{{11 \choose 9}}{{16\choose 9}}$$