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:
- 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
Best Answer
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 $P(b_2|b_1)=\frac{\frac{17}{48}}{\frac58}=\frac{17}{30}$