# Conditional Probability Problem (drawing chips from an urn)

"An urn contains one white chip and a second chip that is equally likely to be white or black. A chip is drawn at random and returned to the urn. Then a second chip is drawn. What is the probability that a white appears on the second draw given that a white appeared on the first draw"?

I keep getting answer 3/4 but it is incorrect. I tried calculating P(A|B) directly; also, I looked at the case where both chips were white then one was black and I averaged those. I always got 3/4.

Personally, I don't even see how the first drawing has any affect on the latter.

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What does your book say is the correct answer? and if this is homework, please add the homework tag. – Dilip Sarwate Mar 21 '12 at 2:46
@Broseph FYI: I removed my wrong comment. – Jeff Mar 21 '12 at 18:29

Before we know what the first drawing is, we have a

• $(1/2)\cdot(1) = 1/2$ chance that both chips in the urn are white, and the first drawing is white.
• $(1/2)\cdot(1)\cdot(1) = 1/2$ chance that both chips in the urn are white, the first drawing is white, and the second drawing is white.
• $(1/2)\cdot(1/2) = 1/4$ chance that the urn has one white and one black, and the first drawing is white.
• $(1/2) \cdot(1/2) \cdot (1/2) = 1/8$ chance that the urn has one white and one black, the first drawing is white, and the second drawing is white.

Therefore the chances that the first drawing is white are $1/2 + 1/4 = 3/4$, and the chances a priori of getting two whites is $1/2 + 1/8 = 5/8$. Therefore the probability of getting a second white given a first white is

$$\frac{\frac{5}{8}}{\frac{3}{4}} = \frac56 .$$

Conceptually, the first drawing doesn't have any effect on the latter after we already know the contents of the urn, but given that the first chip is white, it is more likely that both chips in the urn are white, so we have to weight our probabilities differently.

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Nice, I wasn't too sure why there even was 2 drawings. But now I see that it actually does matter. I just wasn't sure how to write that, so thank you. – The Substitute Mar 21 '12 at 2:58

Conditional probabilities are not a matter of cause and effect; see also this answer I just gave. The question is not whether the first draw has any effect on the second one, but whether your knowledge of the result of the first draw makes a difference for the probabilities you assign to the outcomes of the second one.

If $A$ is the event of drawing white on the second draw and $B$ is the event of drawing white on the first draw, then, since there is an equal chance of $1/2$ each of the second chip being white or black,

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac12\cdot1+\frac12\cdot\frac14}{\frac12\cdot1+\frac12\cdot\frac12}=\frac{1+\frac14}{1+\frac12}=\frac56\;.$$

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Let $D_{1}, D_{2}$ be the events of draw 1 and draw 2 respectively. Use the notation $P_{c=w}(w)$ to denote the probability that a draw is white in the $c=w$ case, where $c$ means the "chip" that is equally likely to be white or black. So $c=b$ would be for the case when it turns out to be black. Then:

$$P(D_{2}=w | D_{1}=w) = \frac{P(D_{2}=w \cap D_{1}=w)}{P(D_{1}=w)}$$ $$= \frac{P_{c=w}(w)P_{c=w}(w)P(c=w) + P_{c=b}(w)P_{c=b}(w)P(c=b)}{P_{c=w}(w)P(c=w) + P_{c=b}(w)P(c=b)}$$ $$= \frac{1\cdot{}1\cdot{}1/2 + 1/2\cdot 1/2 \cdot 1/2}{1\cdot 1/2 + 1/2\cdot 1/2} = \frac{5}{6}.$$

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