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.

  • $\begingroup$ Welcome to MSE. Please edit and use MathJax to properly format math expressions. $\endgroup$ Feb 6, 2020 at 21:31
  • $\begingroup$ 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. $\endgroup$ Feb 6, 2020 at 21:33
  • $\begingroup$ Yes, I was just hoping there was a quicker way than that. $\endgroup$ Feb 6, 2020 at 21:34
  • $\begingroup$ $P(B\ and\ A) = P(B) * P(A\ given\ B)$ 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. $\endgroup$ Feb 6, 2020 at 21:40

2 Answers 2


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

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

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

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

  • $\begingroup$ Your first case should be doubled; not the second $\endgroup$ Feb 6, 2020 at 21:52
  • $\begingroup$ $10 * 9 * 5$ represents "green, green, yellow" or "green, yellow, green" and $10 * 5 * 4$ represents "green, yellow, yellow" $\endgroup$ Feb 6, 2020 at 21:59
  • 1
    $\begingroup$ Oops, I misread the question to be 10 yellow and 5 green $\endgroup$
    – sentheta
    Feb 6, 2020 at 22:03

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