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In a deck of $n$ cards, 30% have white on both sides, 50% have black and white sides, and 20% of cards have black on both sides. The deck is then shuffled fairly and a random card is drawn and set on the table. If the card is black, then what is the probability that the other side is white? (Pitman's Probability, 1.4.8)

Here's what I thought:

We have that P(WW) = 0.3, P(BW) = 0.5, and P(BB) = 0.2. We want to find P(W | B).

P(W | B) = $\frac{P(BW)}{P(B)}$ = $\frac{0.5}{0.5 + 0.2}$ = $\frac{5}{7}$.

However, the solution manual says the answer is $\frac{0.5}{0.5 + 2*0.2}$. According to this, we need to consider P(BB) two times, rather than just once. Why is this? Isn't the probability of a card having black on one side the union of Black & White cards and Black & Black cards? What am I missing here? Thank you.

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  • $\begingroup$ You need to be more specific about what P(W|B) means. I assume it means given that you know one side if black what is the probability that the other side is white. Given one side is black, the only possibilities are BB and BW. BB or BW occur a total of 70% of the time. You get BW 0.5/(0.5+0.2) = 5/7 or approximately 71% of the time when one side is black. This does not agree with the manual but does agree with your calculation. $\endgroup$ – Michael Chernick Sep 9 '17 at 7:01
  • $\begingroup$ My comment would be correct if my interpretation of P(W|B) is correct. $\endgroup$ – Michael Chernick Sep 9 '17 at 7:09
  • $\begingroup$ Another interpretation might be that P(W|B) is the probability that the top side is white given the bottom side is black. Then order matters and {B,W} differs from {W, B}. But we are not told the 50% of cards with one side white and the other black split. If we assume 25% {B,W} and 25% {W, B} then P(W|B) = P(WB)/{P(WB) + P(BB)} =0.25/{0.25+0.2} =25/45 = 55.6%. $\endgroup$ – Michael Chernick Sep 9 '17 at 7:38
  • $\begingroup$ How the orientation of the cards is determined is missing from the problem description. In a deck of standard playing cards, one side of each card is unmarked, but here, both sides can vary. $\endgroup$ – Kodiologist Sep 9 '17 at 19:43
  • $\begingroup$ @MichaelChernick, yeah according to my GSI your 2nd interpretation is the one they're using. The BB cards contribute "twice as much" as the BW cards in terms of black sides, so we need to account for BB cards twice. $\endgroup$ – Max Sep 9 '17 at 22:54
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While the comment from Max answered it, a pedagogical solution is to apply Bayes: $$ P(BW|top=B) = {P(top=B|BW)P(BW)\over P(top=B)} $$ work out the denominator as usual: $$ P(top=B) = P(top=B|WW)P(WW)+P(top=B|BW)P(BW)+P(top=B|BB)P(BB) $$

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