There are 10 boxes containing blue and red balls.

The number of blue balls in the
$n^\text{th}$
box is given by
$B(n) = 2^n$
.

The number of red balls in the
$n^\text{th}$
box is given by
$R(n) = 1024 - B(n)$
.

A box is picked at random, and a ball is chosen randomly from that box. If the ball is blue, and the probability that the $10^\text{th}$ box was picked can be expressed as $\frac ab$ , where $a$ and $b$ are coprime positive integers, find $a+b$ .

More probability questions

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The answer is 1535.

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Let A = probability that the 10th box was picked

Let B = initial probability that a blue ball was chosen.

Using Baye's theorem ,

$P(A|B) = \frac{P(B|A)*P(A)}{P(B)} = \frac{(1)*(1/10)}{(1/10)*\frac{2^1+2^2+...+2^{10}}{1024}} = \frac{1024}{2^1+2^2+...+2^{10}} = \frac{1024}{2046} = \frac{512}{1023}$

And $512 +1023 = \boxed{1535}$