Bayes' Theorem About Drawers and T-Shirts

Question

Drawer 1 has 4 black t-shirts. Drawer 2 has 4 white t-shirts. Drawer 3 has 3 black t-shirts and 1 white t-shirt. Assuming that I drew a black t-shirt from a random drawer, what is the probability that I draw a white t-shirt from the same drawer instead?

Answer 1

Let $B$ be the event that a black shirt is drawn, $W$ be the event that a white shirt is drawn, and $Di$ be the event that drawer $i$ is selected. The question is kind of vague so without knowing more details, I'm assuming that you are drawing without replacement.

We seek $P(W|B) = \frac {P(WB)}{P(B)}.$ Let's compute the numerator and denominator separately and plug them in at the end. For both computations, we can condition on the event of choosing a given drawer, this is the application of Bayes Theorem.

$P(WB) = P(WB|D_1)P(D_1) + P(WB|D_2)P(D_2) + P(WB|D_3)P(D_3)$

As drawers 1 and 2 do not have both white and black shirts, the first and second terms vanish. We are left with

$P(WB) = P(WB|D_3)P(D_3)$ $=$ $\frac{{3 \choose 1}{1 \choose 1}}{4 \choose 2}$$*\frac13 = \frac16$ $P(B) = P(B|D_1)P(D_1) + P(B|D_2)P(D_2) + P(B|D_3)P(D_3)$ and $P(D_i)=\frac13$ for $i = 1,2,3$ Thus, $P(B) = \frac13 *(1 + 0 + \frac34) = \frac7{12}$

Yielding that $P(W|B) = \frac{\frac16}{\frac7{12}} =\frac{2}{7}$