# Draw probability tree for drawing black & white cards (how to use $P(A|B)$)

I have a homework question that reads:

You have a deck of $16$ cards, with:

$4$ cards that are white on both sides;
$7$ cards that are white on one side, black on the other;
$5$ cards that are black on both sides.

The cards are shuffled and randomly flipped. You draw one card from the deck and look only at one side of it.

• (a) Draw a tree diagram with the probabilities of all possible card types. Hint: Your tree should have two levels, one for the color of the top of the card, and one for the color of the bottom card.
• (b) What is the probability that the top of your card is black?
• (c) If the top of your card is black, what is the probability that the bottom is white?
• (d) If the top of your card is white, what is the probability that the bottom is white?

Now, here has been my approach thus far:
first, I figured out how many total white sides and black sides there are:

White sides $= 4 \times 2 + 7 = 15$
Black sides $= 5 \times 2 + 7 = 17$
Total sides $= 32$

Probability for top side being white $= 15/32 = 0.46875$
Probability for top side being black $= 1 - 15/32 = 0.53125$

Now, my problem is there is this $P(A | B)$ formula thing that I'm supposed to use to get the next level ones, but I don't know how to do that, so I skipped the probabilities for the second level in the tree for a bit.

I went on to (c), where I said the probability was $7/12$, or about $0.58333$, because if your top is black, either you have a full black card (of which there are $5$), or you have a mixed card (of which there are $7$). You're looking for the probability of a mixed card, so you have a total of $7$ winning picks out of a total $12$ picks ($7 + 5$). I did a similar thing for (d)

Then I went back and filled in the tree by doing the probability of the top side (say the top is black, which is $0.53125$) times the probability of the bottom for that branch (say bottom is white (top has already been black), so that is $0.58333$). So I went $0.53125 \times 0.58333 = 0.309894$, and did $0.53125 - 0.309894$ to get the probability of a full black card.

My question is, how can I fill out the probability tree by using this $P(A|B)$ thing? I know that the formula for $P(A|B)$ is $P(A \cap B) / P(B)$. The thing is, I was getting an obviously wrong probability.

Let $B$ be the event that the top of the card drawn is black.
Let $A$ be the event that the bottom of the card drawn is white.

$P(A|B) = P(A \cap B) / P(B) = (7/16) / (17/32) = 14/17 = 0.823529\ldots$? what?

I did $7/16$ because that's the percentage of getting a mixed card ($A \cap B$), right? I don't understand

### UPDATE:

Actually, $7/16$ should be $7/12$ right? Because we don't need to have the $4$ full whites in there. Now, $7/12 \times 17/32 = 0.309894$, which is the correct answer I believe. But isn't it $P(A \cap B) / P(B)$?

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The relevant (and very useful) formula is $$\Pr(P|Q)\Pr(Q)=\Pr(P\cap Q),$$ or any of its variants, such as $\Pr(P|Q)=\frac{\Pr(P\cap Q)}{\Pr(Q)}.$ It is in general a good idea to know one of the formulas well, instead of trying to memorize several closely related ones.

We will do formal calculations. It is important to be able to do these. It is even more important to understand the underlying tree, that is, to know what's really going on.

For clarity, I would like to define the events $A$ and $B$ not quite like you did. Let $A$ be the event the card you draw is bicoloured, and let $B$ be the event the upface of the drawn card is black.

The event $B$ can happen in two ways: (i) We drew a card which is bicoloured, and the upface was black or (ii) we drew a two black card and the upface was black.

The probability we draw a bicoloured card is $\frac{7}{16}$. Given that we drew that card, the probability that the upface was black is $\frac{1}{2}$. So the probability of (i) is $\frac{7}{16}\cdot\frac{1}{2}$.

The probability we draw a two-black card is $\frac{5}{16}$. Given that has happened, the probability that the upface is black is $1$. So the probability of (ii) is $\frac{5}{16}\cdot 1$.

It follows that $$\Pr(B)=\frac{7}{16}\cdot\frac{1}{2}+\frac{5}{16}\cdot 1=\frac{17}{32}.$$ If you wish, you can approximate this with a calculator. However, in a several part problem of this type, it is often best to work with the fractions for as long as possible.

Now for (c) all we need $\Pr(A \cap B)$. This is the probability that we drew the bicoloured card and the upface was black. We have already computed this, it is just the probability of (i). (That will often happen in this kind of calculation.) So $\Pr(A\cap B)=\frac{7}{32}$. It follows that $$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}=\frac{\frac{7}{32}}{\frac{17}{32}}=\frac{7}{17}.$$

For (d), let $C$ be the event we draw a two-white card, and let $D$ be the event the upface of the drawn card is white. We want $\Pr(C|D$. The calculation is of the same kind as the previous one. We have $\Pr(C|D)\Pr(D)=\Pr(C\cap D)$.

For $\Pr(D)$, follow the tree. We get upface white if we drew a two-white card (probability $\frac{4}{16}$) or we drew a bicoloured card and got upface white (probability $\frac{7}{16}\cdot\frac{1}{2}$), for a total of $\frac{15}{32}$. We need not have done the above calculation, since clearly $\Pr(D)=1-\Pr(B)=1-\frac{17}{32}$.

During the previous calculation, we found $\Pr(C\cap D)$: it is $\frac{4}{16}$. So $$\Pr(C|D)=\frac{\frac{4}{16}}{\frac{15}{32}}=\frac{8}{15}.$$

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Thank you so much for taking the time to develop this very thorough explanation, you're a good person :) –  user13327 Sep 2 '12 at 15:46

When shuffling and flipping at random, each of the following 32 configurations is equally likely: $$8: (W,W), \quad 7: (W,B), \quad 7:(B,W), \quad 10:(B,B)$$ where the first letter denotes the color of the top side of the card, and the second denotes the color of the bottom side. The top of the first card is black in $10+7$ cases out of $32$. Thus, for b), the probability is $\frac{17}{32}$. Out of these $17$ cards, $7$ will have white bottom side, and $10$ will have black bottom side. Continuing in this manner produces the following tree diagram:

Now answers to remaining questions can be easily read off from this diagram.

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Thank you very much Sasha! I wish I could mark two as the correct answer, as I used both of your explanations to finally break through my mental block on this one. Thanks! –  user13327 Sep 2 '12 at 15:45

For (c), you have 17 black sides, of which you have randomly selected one. The other sides are 10 black, 7 white. So given that the top side is black, the probability that the bottom is white is $\frac 7{17}$. The argument for (d) is similar.

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Ah! Oops, thank you! –  user13327 Sep 2 '12 at 15:50