Conditional Probability- With replacement

Question

Say a bag contains 2 white balls and 3 black balls. 3 balls are drawn randomly with replacement. a. Find the P(1st is white | 2nd is black) b. Find the P(1st is white | at least 1 black ball is drawn)

For a. I found the P(W)=2/5 and the P(B)=3/5, then I multiplied those together to get 6/25. Then I used the formula for condition probability P(WB)/P(2nd is B)= (6/25)/(3/5)= 2/5, is this correct?

And then for b. I got stuck at P(at least 1 ball is B), I know I should do 1-P(only white balls are drawn) and to find that I thought about doing 5C2/5C3. Is this the correct way to find the P(at least 1 black ball is drawn?

Answer 1

Your answer for (a) is correct.

For (b) the probability is as you suggest $1-(\frac{2}{5})^3=\frac{117}{125}$.

On the top line you have the probability that the first ball is white and then the next two balls are not both white. This is $(\frac{2}{5})(1-(\frac{2}{5})^2)=\frac{42}{125}$.

So the conditional probability is $\frac{42}{117}.$

Answer 2

(a) You're right. Notice that the multiplication you did is justified because the events B and W are independent (because the balls are drawn with replacement), and the $P(B)$ cancels out in the end. In general, such a calculation shows that if A and B are independent events then $P(A|B)=P(A)$, which matches our intuition about what "independence" is.

(b) You're right that you want to find $1-P(A)$ where $A$ is the event that only white balls are drawn. Remember that we're drawing with replacement, so each draw is independent. Therefore, $P(A)=P(\text{all balls white})=P(\text{first ball white})P(\text{second ball white})P(\text{third ball white})=\frac23\frac23\frac23.$

Answer 3

Let us denote $B_1, B_2, B_3$ to represent the independent events of drawing a black ball in the indicated draw; and likewise their corresponding complements shall be $W_1, W_2, W_3$.

(a) No need to multiply then divide by the same thing.   It is simply that $\mathsf P(W_1\mid B_2)= \mathsf P(W_1)$ since these events are independent (vis "with replacement"), so the result is as you had: $\mathsf P(W_1\mid B_2)= 2/5$

$$\mathsf P(W_1\mid B_2)=\dfrac{\mathsf P(W_1)\,\require{cancel}\cancel{\mathsf P(B_2)}}{\cancel{\mathsf P(B_2)}}$$

(b) "At least one is black" is the event $B_1\cup B_2\cup B_3$, and this will be the complement of $W_1\cap W_2\cap W_3$, as you reasoned. However, there will be no binomial coefficients involved in this evaluation, as the draws are made with replacement.

The definition of conditional probability, and the independence of the draws, leads to:

$$\mathsf P(W_1\mid B_1\cup B_2\cup B_3)=\dfrac{\mathsf P(W_1\cap( B_2\cup B_3) )}{\mathsf P(B_1\cup B_2\cup B_3)}=\dfrac{\mathsf P(W_1)\,(1-\mathsf P(W_2)\,\mathsf P(W_3))}{1-\mathsf P(W_1)\,\mathsf P(W_2)\,\mathsf P(W_3)}$$