Probability of Picking Socks finding the intersection of two dependent events

Question

A box is filled with 15 different socks: 10 green and 5 yellow. Three socks are picked from the box (leaving 12 in the box). Let $x$ be the random variable describing the number of green socks that are selected. Let events $A$ and $B$ be described as follows:

$A$ = the event that not all three socks are green.

$B$ = the event where the first sock selected is green.

Find $P(A)$, $P(B)$, and $P(A\text{ and }B)$. Show any appropriate $nCr $ expressions.

I know for this problem that $P(A) = 1 - (\frac{10}{15} \times \frac{9}{14} \times \frac{8}{13})$ and $P(B) = \frac{10}{15}$, but I'm not sure how to find $P(A\text{ and }B)$. I know these events are dependent.

Answer 1

$$P(A\text{ and }B) = P(B) \times P(A\ |\ B)$$

$$P(B) = \frac{10}{15}\tag{as you have shown}$$

The probability of $A$ given $B$ is equal to the probability of $A$ occurring after $B$ has occurred (i.e. after a green sock has been removed). Once a green sock has been removed, the probability of not all three socks being green is the probability of getting 2 non-green socks where 9 of the 14 socks are green.

$$P(A\ |\ B) = (1-\frac{9}{14}\times\frac{8}{13})\tag{remove first choice of green}$$

$$P(A\text{ and }B) = \frac{10}{15} \times (1-\frac{9}{14}\times\frac{8}{13}) = \frac{110}{273}$$

Answer 2

Separate $P(A\ \Bbb{and}\ B)$ into 2 cases. First case where 1 green sock is selected. Second case where 2 green socks are selected. Note that there are 2 possibilities for the second case, the last sock is green or the last sock is yellow.

Case 1 : No. of events $=10\times 5\times 4=200$

Case 2 : No. of events $=2\times(10\times 9\times 5)=900$

$$P(A\ \Bbb{and}\ B)=\frac{200+900}{15\times 14\times 13}=\frac{110}{273}$$