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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?

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up vote 2 down vote accepted

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.

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The following is imprecise, but may help in the visualization. Imagine that Ann does the experiment $2(7700)$ times. We picked this strange number to make the later arithmetic simple.

Then "about" $7700$ times she chooses the box with $3$ orange and $4$ black, and "about" $7700$ times she chooses the box with $5$ orange and $6$ black.

Out of the about $7700$ times she chooses the first box, she gets orange "about" $\frac{3}{7}(7700)$ times, that is, $3300$ times. And out of the roughly $7700$ times she chooses the second box, she gets orange about $\frac{5}{11}(7700)$ times, so roughly $3500$ times.

So she got orange about $6800$ times. And they came from the second box about $3500$ times. So we would expect that the probability the ball came from the second box given that it is orange should be $\dfrac{3500}{6800}$.

It is safer to use the machinery of conditional probabilities. But the above calculation may help give an informal idea of what is going on. In essence we are resricting the sample space to those situations where we got an orange ball.

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