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An urn contains 10 white balls and 5 blue balls. Draws are made repeatedly from the urn as follows. On each draw, a ball is drawn and its color noted; then it is replaced in the urn along with 3 more balls of its color.

Find the chance that the first ball drawn is blue, given that the second ball drawn is blue.

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in your answer why would you say that that the first ball drawn is white, which occurs with probability 2/3, how to find 1/3 or 2/3? Thanks –  user73422 Apr 19 '13 at 19:47
    
With all due respect to the helpful people here, I do want to make the point that, in all probability, user73408 is a student of the edX course Stat2.2X, and that he or she has broken the honor code of that course by asking someone to solve this problem before the submission deadline. I do not profess to have the "right answer" to this issue, nor am I asking people not to be helpful. I just wanted to add some pertinent information. –  user73598 Apr 21 '13 at 5:32

1 Answer 1

Such problems are usually introduced as applications of Bayes’ theorem, but this isn’t necessary.

  • If the first ball drawn is white, which occurs with probability $\frac23$, the urn will contain $13$ white and $5$ blue balls for the second draw, so the probability of drawing a blue ball will be $\frac5{18}$.

  • If the first ball drawn is blue, which occurs with probability $\frac13$, the urn will contain $10$ white and $8$ blue balls for the second draw, so the probability of drawing a blue ball will be $\frac8{18}$.

The overall probability of drawing a blue ball on the second draw is therefore

$$\frac23\cdot\frac5{18}+\frac13\cdot\frac8{18}\;.\tag{1}$$

What fraction of this probability corresponds to the case in which the first ball drawn is blue? That’s the conditional probability that the first ball drawn is blue, given that the second is blue.

Added: We see from $(1)$ that the overall probability of getting a blue ball on the second draw is $\frac{18}{54}=\frac13$. It’s given that the second ball drawn was blue, so we know that we’re in this situation. How did we get there?

  • The first ball drawn might have been white; that case contributes $\frac23\cdot\frac5{18}=\frac{10}{54}=\frac5{27}$ to the overall probability of getting a blue ball on the second draw.

  • The first ball drawn might have been blue; that case contributes $\frac13\cdot\frac8{18}=\frac4{27}$ to the overall probability of getting a blue ball on the second draw.

Imagine that we play this game $27$ times, and the results just happen to match the probabilities perfectly. Then we’ll draw a white ball the first time and a blue ball the second time in $5$ of those games, and we’ll draw blue balls both times in $4$ of the games. (Note that we get a blue ball on the second draw $9$ times, i.e., in $\frac13$ of the $27$ games, corresponding to the probability of $\frac13$ from $(1)$.) So in $4$ of the $9$ games in which we draw a blue ball the second time we had already drawn a blue ball the first time; the probability that the first ball was blue, given that the second is blue, must therefore be $\frac49$.

But we didn’t need to imagine these $27$ plays that match the probabilities exactly: we could simply have answered my original question. The second term of $(1)$ is the part that corresponds to the case in which the first ball is blue, and it contributes

$$\frac{\frac13\cdot\frac8{18}}{\frac23\cdot\frac5{18}+\frac13\cdot\frac8{18}}=\frac{\frac4{27}}{\frac13}=\frac49$$

of the total probability of getting a blue ball on the second draw. The case of getting a white ball on the first draw contributes the other $\frac59$ of the total. Thus, if you get a blue ball on the second draw, about $\frac49$ of the time you had a blue ball on the first draw, and about $\frac59$ of the time you had a white ball.

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I am sorry, the answer of 18/54 is Incorrect. –  user73408 Apr 19 '13 at 19:21
    
@user73408: I did not say that $\frac{18}{54}$ was the answser to your question; it isn’t. It’s a number that you need in order to get the answer to your question. The way to get that answer is given in the last two sentences of my answer. –  Brian M. Scott Apr 19 '13 at 19:24
    
Oh, OK, sorry, can i have a detail explaination.... how to find the final result. Thanks –  user73408 Apr 19 '13 at 19:32
    
@BrianM.Scott Although you claim that Bayes' theorem is not necessary, you do in effect use Bayes' theorem in conjunction with the law of total probability (that is, you use $$P(B\mid A) = \frac{P(A\mid B)P(B)}{P(A\mid B)P(B)+P(A\mid B^c)P(B^c)}$$ instead of the slicker version $$P(B\mid A)=\frac{P(A\mid B)P(B)}{P(A)}$$ that is usually found in textbooks and befuddles countless beginners in probability) without calling it Bayes' theorem or formula. –  Dilip Sarwate Apr 19 '13 at 19:51
    
@Dilip: Oh, I know that. My point is that the necessary idea is perfectly straightforward and can be reasoned through without appealing to the magical, mysterious formula. –  Brian M. Scott Apr 19 '13 at 19:53

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