What's the real solution to this exercise?

Question

The exercise says like this

There are 3 boxes containing balls:

• The first has 3 white and 2 black
• The second has 2 white and 3 black
• The third has 3 white and 3 black

Only one ball is chosen from each box. If two of the chosen balls are white:

• What is the probability that one of the balls was chosen from the first box?

The result is 15/19

• So, what's the real solution to this exercise?

The data I gather so far are

• Ci: "The ball is of the box i" such that i = 1 or 2 or 3
• B: "The ball is white"

• N: "The ball is black"

  P(B|C1) = 3/5, P(N|C1) = 2/5

  P(B|C2) = 2/5, P(N|C2) = 3/5

  P(N|C3) = 1/2, P(B|C3) = 1/2

  P(Ci) = 1/3

  P(B|C1unionC2unionC3) = 2/3, Or should it be an intersection between the three?

  P(B) = P(B|C1)*P(C1) + P(B|C2)*P(C2) + P(B|C3)*P(C3)

  P(B) = 1/2

  P(C1|B) = P(B|C1)*P(C1) / P(B) = 2/5

Answer 1

There are three scenarios that yield exactly two white balls:

$$P(W,W,B)=\frac 35 \times \frac 25\times \frac 36= \frac 3{25}$$

$$P(W,B,W)=\frac 35 \times \frac 35\times \frac 36= \frac 9{50}$$

$$P(B,W,W)=\frac 25 \times \frac 25\times \frac 36= \frac 2{25}$$

The answer you seek is $$\frac {P(W,W,B)+P(W,B,W)}{P(W,W,B)+P(W,B,W)+P(B,W,W)}=\frac {6+9}{6+9+4}=\frac {15}{19}$$

Answer 2

This is an application of Baye's theorem.

Let $A$ be the event: exactly two white balls appear.

Let $B$ be the event: A white ball is extracted from the first bin.

We want to calculate $P(B|A)$.

By Baye's theorem it's equal to $\frac{P(A|B)P(B)}{P(A)}=\frac{(2/5\times 1/2+ 3/5\times 1/2)(3/5)}{P(A)}=\frac{3}{10P(A)}$.

So we need to calculate $P(A)$.

There are $3$ ways in which exactly $3$ balls can be extracted, we can add the probabilities to get: $3/5\times2/5\times 1/2+ 2/5\times 2/5\times 1/2+ 3/5\times 3/5\times 1/2=\frac{19}{50}$

So the final answer is $\frac{150}{190}=\frac{15}{19}$