There are four boxes. Each contains 2 balls. The first box has a red and a white ball in it. The remaining three boxes each have two white balls in them.

A ball is picked at random from box 1 and put in box 2.

Then a ball is picked at random from box 2 and put into box 3.

Then a ball is picked at random from box 3 and put into box 4.

Finally, a ball is picked from box 4.

The probability that the ball picked from box 2 is red, given that the final ball picked from box 4 is white can be written as $\dfrac ab$ , where $a$ and $b$ are coprime positive integers, find $a+b$ .

More probability questions

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The answer is 61.

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Let R2 = probability that a red ball was picked from the second box

Let W4 = probability that a white ball was picked out from the fourth box

Using Baye's theorem ,

$P(R2|W4) = \frac{P(W4|R2)*P(R2)}{P(W4)} = \frac{(8/9)*(1/6)}{53/54} = \frac{8}{53}$

And $8 +53 = \boxed{61}$