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I'm stuck with this question (sorry if it's ridiculously simple, I'm new to this): A person keeps his socks in two drawers. The first drawer contains 8 socks of which 4 are white. The second drawer contains 10 socks of which 8 are white. He chooses one drawer randomly and picks two socks from that drawer. Find the probability that both socks are white. I've calculated the probability for getting 2 white socks from drawer 1 (3/14) or from drawer 2 (14/45) (using the formula for B, given that A has happened). But what now??

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  • $\begingroup$ Probability of drawing 2 white socks from first drawer is the probability of drawing a white sock (B_1), given that a white sock has already been drawn (A_1). P(A_1∩B_1 )=P(A_1 )P(B_1│A_1 ) Probability of the first sock being white is P(A_1 )=4/8=1/2 Probability of second sock drawn being white, given that first is white is P(B_1│A_1 )=3/7 Therefore P(A_1∩B_1 )=1/2∙3/7=3/14 $\endgroup$ – Annukka Sep 17 '18 at 18:54
  • $\begingroup$ Yes, sorry. I misread and thought there were $12$ socks in the first drawer. To your question: there is a $\frac 12$ chance of using the first drawer, and the same for the second. Does that clarify the point? $\endgroup$ – lulu Sep 17 '18 at 18:55
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    $\begingroup$ Note: for the second drawer, the probability of drawing two white socks is $\frac 8{10}\times \frac 79=\frac {28}{45}$ unless I am misreading again. $\endgroup$ – lulu Sep 17 '18 at 18:56
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Let $D_1$ be that our person picked the first drawer. Let $D_2$ be that our person picked the second drawer. Let $W_1$ be that the first sock selected is white. Let $W_2$ be that the second sock selected is white.

Note that $D_1\cap D_2=\emptyset$ and that $D_1\cup D_2=\Omega$

We are tasked with calculating $Pr(W_1\cap W_2)$, that both socks selected are white.

By total probability and by multiplication principle of probability we have:

$Pr(W_1\cap W_2)=Pr(D_1)Pr(W_1\mid D_1)Pr(W_2\mid D_1\cap W_1) + Pr(D_2)Pr(W_1\mid D_2)Pr(W_2\mid D_2\cap W_1)$

$=\frac{1}{2}\cdot \frac{4}{8}\cdot \frac{3}{7} + \frac{1}{2}\cdot\frac{8}{10}\cdot\frac{7}{9}$

This may all be visualized through the use of a tree diagram.

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  • $\begingroup$ Thanks for your answer - getting my head around this properly might take a minute or two! In the meanwhile, was I going on a wild goose chase with working out the two probabilities separately to begin with? $\endgroup$ – Annukka Sep 17 '18 at 19:00
  • $\begingroup$ No. What you did was the same as what I did, I just phrased the final answer all at once. Your separate probabilities are simply the $\frac{4}{8}\cdot\frac{3}{7}$ and $\frac{8}{10}\cdot\frac{7}{9}$ parts respectively. The only step you were missing was multiplying each by $\frac{1}{2}$ and adding to represent the action of selecting which drawer to draw from first. $\endgroup$ – JMoravitz Sep 17 '18 at 19:01

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