# Probability of Picking Socks finding the intersection of two dependent events

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.

• Welcome to MSE. Please edit and use MathJax to properly format math expressions. Feb 6, 2020 at 21:31
• One way of going about this problem would be to draw a tree of the probabilities and add together the probabilities where A & B occur. Feb 6, 2020 at 21:33
• Yes, I was just hoping there was a quicker way than that. Feb 6, 2020 at 21:34
• $P(B\ and\ A) = P(B) * P(A\ given\ B)$ So you could multiply your result from $P(B)$ by the probability of $B$ occurring after having removed a green sock. I have not tested this out, but should work. Feb 6, 2020 at 21:40

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

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

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.

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

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

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 $$=2\times(10\times 9\times 5)=900$$

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

• Your first case should be doubled; not the second Feb 6, 2020 at 21:52
• $10 * 9 * 5$ represents "green, green, yellow" or "green, yellow, green" and $10 * 5 * 4$ represents "green, yellow, yellow" Feb 6, 2020 at 21:59
• Oops, I misread the question to be 10 yellow and 5 green Feb 6, 2020 at 22:03