Probability of picking balls out of bins

Question

Question: You have two bins with four different balls in each bin.

Bin A: 2 White Balls and 2 Black Balls

Bin B: 3 Black Balls and 1 White ball

You cannot tell which bin contains what balls. Given that you chose a bin at random, and from that bin you picked a Black ball, what is the probability that you will pick another Black ball. Assume that on the second turn you also choose another bin at random, but now one is minus one black ball.

My Question is:

1. Is it possible to get a single probability value for this question? I made this question up on the fly as a way to solidify my understanding of probability, but I'm unsure if it can be answered by a single value.

To extrapolate, I can get probabilities for P("Chose Bag A"|"drew a black Ball") and P("Chose Bag B"| "drew a back ball") easy enough with Bayes rule, but I'm not sure if it is possible to get an answer for P("drew a black ball"|"drew a black ball").

I'm looking for some guidance here as to what I'm doing wrong, if I did frame the question wrong, or if it is possible, how to do the question.

Thank you

Answer 1

Let $b_1$ be the event drawing a black ball the first draw; and $b_2$ the event of drawing a black ball the second draw.

We want $P(b_2|b_1)$. That is, we want $\frac{P(b_2\cap b_1)}{P(b_1)}$.

We have $P(b_1)=\frac12\cdot\frac12+\frac12\cdot\frac34=\frac58$.

To figure out $P(b_1\cap b_2)$ we note that there are four ways this can happen:

(i) Bin A, Black, Bin A, Black, with probability $\frac12\cdot\frac12\cdot\frac12\cdot\frac13=\frac1{24}$

(ii) Bin A, Black, Bin B, Black, with probability $\frac12\cdot\frac12\cdot\frac12\cdot\frac34=\frac3{32}$

(iii) Bin B, Black, Bin A, Black, with probability $\frac12\cdot\frac34\cdot\frac12\cdot\frac12=\frac3{32}$

(iv) Bin B, Black, Bin B, Black, with probability $\frac12\cdot\frac34\cdot\frac12\cdot\frac23=\frac18$

So $P(b_1\cap b_2)=\frac{1}{24}+\frac3{32}+\frac3{32}+\frac18=\frac{34}{96}=\frac{17}{48}$

Then the overall conditional probability $P(b_2|b_1)=\frac{\frac{17}{48}}{\frac58}=\frac{17}{30}$

Answer 2

One thing I would change in the way you phrased the problem is to be sure you distinguish the event of getting black ball on the first draw from the event of getting a black ball on the second draw. These are two completely separate (though not independent) events.

Clearer notation could help. You have to distinguish such things as "selected bin $A$ for the first ball drawn" and "selected bin $A$ for the second ball drawn." It quickly gets cumbersome to write the probability formulas if you have to spell out the events in so many words every time. So how about defining some symbols for this problem like this:

For $n = 1$ or $n = 2,$ let $A_n$ be the event that you select bin $A$ to draw the $n$th ball from, $B_n$ be the event that you select bin $B$ to draw the $n$th ball from, and $C_n$ be the event that the $n$th ball you draw is black.

You want to compute $P(C_2 \mid C_1).$ This is a perfectly reasonable question.

You know $P(A_1).$ You can compute $P(C_1 \mid A_1).$ You should also be able to compute $P(C_2 \mid A_1 \cap C_1).$ From these three things and the chain rule of probability, you can compute $P(A_1 \cap C_1 \cap C_2).$

By a similar procedure you can compute $P(B_1 \cap C_1 \cap C_2).$

But $A_1 \cap C_1 \cap C_2$ and $B_1 \cap C_1 \cap C_2$ are disjoint events, and $$(A_1 \cap C_1 \cap C_2) \cup (B_1 \cap C_1 \cap C_2) = (A_1 \cup B_1) \cap C_1 \cap C_2 = C_1 \cap C_2.$$ Using these facts, you can compute $P(C_1 \cap C_2).$

But $P(C_2 \cap C_1) = P(C_2 \mid C_1) P(C_1).$ Compute $P(C_1),$ then solve the equation for $P(C_2 \mid C_1).$