I am having difficulty understanding the answer to this problem due to the conditional replacement. My initial thought is that I could draw a tree diagram and use the multiplication rule for independent events to find probability of white being drawn but I'm confused by not knowing the first selection and the conditional nature of the problem?

An urn contains three blue and seven white chips. A chip is selected at random. If the color of the chip selected is white, it is replaced and two more white chips are added to the urn. However, if the chip drawn is blue, it is not replaced and no additional chips are put in the urn. A second chip is drawn after the first (problem does not specify which color chip was drawn first). What is the probability that it is white? Supposing that we are given that the chip selected for the second time is white, what is the probability that the first chip selected is blue?

  • $\begingroup$ "...for independent events..." Be careful about using that word. Yes, you will use the multiplication rule here however they are not independent. $\endgroup$
    – JMoravitz
    Jan 15, 2021 at 22:33

2 Answers 2


Let $A$ be the event that a white was drawn on the first pull. Let $B$ be the event that a white was pulled on the second pull.

You are asked to find $\Pr(B)$ and you are asked to find $\Pr(A^c\mid B)$

To do this, note that it is clear how to calculate $\Pr(A)$ as this is just going to be the ratio of white chips compared to total number of chips in the initial configuration of the urn. It is also apparent how to calculate $\Pr(B\mid A)$ and $\Pr(B\mid A^c)$ as being the ratio of white chips to all chips in the configurations of the urn had we picked a white on the first pull or not respectively.

$\Pr(A)=\frac{7}{10},\Pr(B\mid A)=\frac{9}{12},\Pr(B\mid A^c)=\frac{7}{9}$

Similarly, we can calculate the opposites of these events by subtracting away from $1$

$\Pr(A^c)=\frac{3}{10},\Pr(B^c\mid A)=\frac{3}{12},\dots$

Next piece of information we want is to calculate $\Pr(B)$. For this, recall the Law of Total Probability and multiplication principle

$\Pr(B)=\Pr(A\cap B)+\Pr(A^c\cap B) = \Pr(A)\Pr(B\mid A)+\Pr(A^c)\Pr(B\mid A^c)$

That should get you far enough to have $\Pr(B)$ calculated.

Now, to calculate $\Pr(A^c\mid B)$, apply Bayes' Theorem which is really just repeated application of the definition of conditional probability to get:

$\Pr(A^c\mid B)=\dfrac{\Pr(A^c\cap B)}{\Pr(B)}=\dfrac{\Pr(B\mid A^c)\Pr(A^c)}{\Pr(B)}$

Each value of which we have figured out earlier in the problem.


As @JMoravitz says you should forget about 'independent events' ideas. You can still multiply along branches of a probability tree and that is a good method to use for this problem.

From the initial scenario of $3$B,$7$W you have a branch leading to $2$B,$7$W and one leading to $3$B,$9$W.

Is that sufficient info. for you to be able to draw the tree and insert probabilities?

For the first part, can you now see that the probability Chip 2 is white is $$3/10\times 7/9+7/10\times 9/12?$$

  • $\begingroup$ I see how that works using the probability diagram. Is this a technique that allows me to easily find P(B) using the law of total probability and multiplication principle (as described in the answer by @JMoravitz) without drawing out the equations. That answer makes sense to me. And is this only applicable for dependent events? $\endgroup$
    – keebler
    Jan 15, 2021 at 23:47
  • $\begingroup$ It is a generally applicable method. The probabilities you put on the branches encompass all issues re. dependence because they are the probabilities of taking that branch given that you are at the previous vertex. $\endgroup$
    – user502266
    Jan 15, 2021 at 23:54
  • $\begingroup$ P.S. To develop a good understanding of probability you should learn to use both methods fluently. In practice, I think students find the tree diagram easier to understand initially. Personally, I often use tree diagrams to explain ideas to students but I would use the algebraic method otherwise, for example because it's quicker not to have to draw a tree! $\endgroup$
    – user502266
    Jan 16, 2021 at 0:01
  • $\begingroup$ It sounds like there is a bit of confusion here... the "tree method" and the "algebraic method" are the same here. The tree is merely a way of visually organizing the information. The math and calculations involved are exactly the same whether you drew the tree or not. At its best, the tree helps remind how to perform the calculations and helps keep numbers straight in your head so you don't confuse which probability was associated with which event. At its worst though, the tree may require more time as you might also fill out unnecessary information for the desired calculation. $\endgroup$
    – JMoravitz
    Jan 16, 2021 at 19:35

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