# Thread: Probability Quests

1. ## Probability Quests

1. A wine collection consists of $4$ bottles of cabernet and $7$ bottles of zinfandel. Randomly selecting $6$ bottles, find the following:

(a) $P \{\text{all cabernets are selected} \} = \frac{\binom{6}{4}}{\binom{11}{6}}$

(b) $P \{\text{at least 1 is a cabernet} \} = \frac{\binom{6}{1} + \binom{6}{2} + \binom{6}{3} + \binom{6}{4} + \binom{6}{5} + \binom{6}{6}}{\binom{11}{6}}$

Are these correct?

2. Try on these.
a) $\frac {{7 \choose 2}} {{11 \choose 6}}$

b) $1 - \frac {{7 \choose 6}} {{11 \choose 6}}$

Can you explain these to us?

3. Originally Posted by heathrowjohnny
1. A wine collection consists of $4$ bottles of cabernet and $7$ bottles of zinfandel. Randomly selecting $6$ bottles, find the following:

(a) $P \{\text{all cabernets are selected} \} = \frac{\binom{6}{4}}{\binom{11}{6}}$

(b) $P \{\text{at least 1 is a cabernet} \} = \frac{\binom{6}{1} + \binom{6}{2} + \binom{6}{3} + \binom{6}{4} + \binom{6}{5} + \binom{6}{6}}{\binom{11}{6}}$

Are these correct?
No.

(a) $P \{\text{all cabernets are selected} \} = \frac{\binom{4}{4} \binom{7}{2}}{\binom{11}{6}} = \frac{\binom{7}{2}}{\binom{11}{6}}$.

(b) The hard way:
$P \{\text{1 is a cabernet} \} = \frac{\binom{4}{1} \binom{7}{5}}{\binom{11}{6}}$.

$P \{\text{2 are cabernet} \} = \frac{\binom{4}{2} \binom{7}{4}}{\binom{11}{6}}$.

$P \{\text{3 are cabernet} \} = \frac{\binom{4}{3} \binom{7}{3}}{\binom{11}{6}}$.

$P \{\text{4 are cabernet} \} = \frac{\binom{4}{4} \binom{7}{2}}{\binom{11}{6}} = \frac{\binom{7}{2}}{\binom{11}{6}}$.

Now add them together.

(b) The easy way:

P{At least 1 is a cabernet} = 1 - P{0 are cabernet}.

$P \{\text{0 are cabernet} \} = \frac{\binom{4}{0} \binom{7}{6}}{\binom{11}{6}} = \frac{\binom{7}{6}}{\binom{11}{6}}$.

4. (a) If you have 6 bottles and you choose 4 cabarets, you are left with 2 extra bottles. So this is the same as $\binom{7}{2}$. How many ways you can choose 2 bottles from the remaining 7 zinfandel.

(b) Took the complement.