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,
\(P(B/D) = \frac {P(B)\times P(D/B)}{P(A)\times P(D/A)+P(B)\times P(D/B) +P(C)\times P(D/C)}\)
= \(\frac {\frac{1}{3}\times \frac{5}{10}}{\frac{1}{3}\times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}+ \frac{1}{3}\times \frac{6}{10}}\) = \(\frac {\frac{1}{3}\times \frac{5}{10}}{\frac{1}{3}\times \frac{15}{10}}\)= \(\frac {1}{3}\)