Three boxes contain 6 white, 4 blue; 5 white, 5 blue and 4 white, 6 bl

1.	Three boxes contain 6 white, 4 blue; 5 white, 5 blue and 4 white, 6 blue balls respectively. One of the box is selected at random and a ball is drawn from it. If the ball drawn is blue, find the probability that it is from the second box.
Answer» 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, \(P(B/D) = \frac {P(B)\times P(D/B)}{P(A)\times P(D/A)+P(B)\times P(D/B) +P(C)\times P(D/C)}\) = \(\frac {\frac{1}{3}\times \frac{5}{10}}{\frac{1}{3}\times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}+ \frac{1}{3}\times \frac{6}{10}}\) = \(\frac {\frac{1}{3}\times \frac{5}{10}}{\frac{1}{3}\times \frac{15}{10}}\)= \(\frac {1}{3}\)

Three boxes contain 6 white, 4 blue; 5 white, 5 blue and 4 white, 6 blue balls respectively. One of the box is selected at random and a ball is drawn from it. If the ball drawn is blue, find the probability that it is from the second box.

Answer»

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,

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

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