Let A, B, C, D be the events defined as:

A: Selecting the first box

B: Selecting the second box

C: Selecting the third box

D: Event of drawing a blue ball.

Since there are three boxes and each box has an equally likely chance of selection, P(A) = P(B) = P(C) = \(\frac{1}{3}\)

♦ If the first box is chosen, i.e., A has already occurred, then

Probability of drawing a blue ball from A = \(\frac{4}{10}\)⇒ P(D/A) =\(\frac{4}{10}\)

♦ If the second box is chosen, i.e., B has already occurred, then

Probability of drawing a blue ball from B =\(\frac{5}{10}\) ⇒ P(D/B) = \(\frac{5}{10}\)

Similarly P(D/C) = \(\frac{6}{10}\)

Now we are required to find the probability (B/D), i.e., given that the ball drawn is blue, we need to find the probability that it is drawn from the second box.

**By Baye’s Theorem,**

### **\(P(B/D) = \frac {P(B)\times P(D/B)}{P(A)\times P(D/A)+P(B)\times P(D/B) +P(C)\times P(D/C)}\)**

## = **\(\frac {\frac{1}{3}\times \frac{5}{10}}{\frac{1}{3}\times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}+ \frac{1}{3}\times \frac{6}{10}}\)** = **\(\frac {\frac{1}{3}\times \frac{5}{10}}{\frac{1}{3}\times \frac{15}{10}}\)**= **\(\frac {1}{3}\)**