Probability with conditional replacements

Question

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?

Answer 1

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 $$3/10\times 7/9+7/10\times 9/12?$$

Answer 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.