Probability – picking up from a box

probability

I have the following question, but I fail to get the right answer. There are two boxes – box $A$ and box $B$. Box $A$ has $5$ red balls and $3$ blue ones. Box $B$ has $6$ red balls and $2$ blue ones. If one randomly chooses one of the boxes and picks up $3$ balls randomly without returning them, I wish to find the probability of the third one being red considering the first two were blue. What I did was:
$$P=P\left(\left(\text{third red}\mid\text{first two blue}\right)\mid\text{box A}\right)P\left(\text{box A}\right)+P\left(\left(\text{third red}\mid\text{first two blue}\right)\mid\text{box B}\right)P\left(\text{box B}\right)=\frac{5}{6}\cdot\frac{1}{2}+\frac{6}{6}\cdot\frac{1}{2}=\frac{11}{12}$$
But the answer seems to be $\frac{7}{8}$. What have I done wrong? I fail to see my mistake (it could be that the answers sheet contains a mistake I guess, but I tend to think I did one).

Best Answer

You have to attack the problem in stages.

First, you have to calculate $f(a), f(b)$ where

$f(a) =$ probability that balls are being drawn from box A and

$f(b) =$ probability that balls are being drawn from box B.

Naturally, this will involve Bayes Theorem, and you should expect $f(a) + f(b) = 1.$

Once you do that, you then have to calculate

$g(a) =$ chance that next ball is red, given that balls are being taken from box A and

$g(b) =$ chance that next ball is red, given that balls are being taken from box B

Then, the final computation will be

$$\left[f(a) \times g(a)\right] + \left[f(b) \times g(b)\right].$$

$f(a) = \frac{R}{S}$

where $R = \frac{1}{2} \left[\frac{3}{8} \times \frac{2}{7}\right]$

and $S = \left\{\frac{1}{2} \left[\frac{3}{8} \times \frac{2}{7}\right]\right\} + \left\{\frac{1}{2} \left[\frac{2}{8} \times \frac{1}{7}\right]\right\}.$

This works out to $f(a) = (3/4), f(b) = (1/4).$

Then, you also have that $g(a) = (5/6)$ and $g(b) = 1$.

The numbers immediately above are explained by reasoning that if the balls are being drawn from box A, there are 6 balls left, 5 of which are red. If the balls are being drawn from box B, then there are 6 balls left, all of which are red.

Thus, the final computation is

$$\left[(3/4) \times (5/6)\right] + \left[(1/4) \times (1)\right] = (21/24) = (7/8).$$

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