Suppose that Ann selects a ball by first picking one of two boxes at random and then selecting a ball from this box. The first box contains three orange balls and four black balls, and the second box contains five orange balls and six black balls. What is the probability that Ann picked a ball from the second box if she has selected an orange ball?
How can I think about the way this problem works? Can someone draw a diagram or something to help me visualize the probabilities we're using?
Best Answer
Ann's probability of choosing the first box and then an orange ball from it is $\frac{1}{2}\times\frac{3}{7}=\frac{3}{14}$. Her probability of choosing the second box and then an orange ball from it is $\frac{1}{2}\times\frac{5}{11}=\frac{5}{22}$. Given that she did, in fact, wind up with an orange ball, one of these two events must have happened; the conditional probability of each is $$ \frac{\frac{3}{14}}{\frac{3}{14}+\frac{5}{22}} = \frac{33}{68} \text{(box 1);}\qquad \frac{\frac{5}{22}}{\frac{3}{14}+\frac{5}{22}} = \frac{35}{68} \text{(box 2).} $$ Since the second box is slightly "oranger" than the first, it makes sense that the second probability should be slightly higher.