Bag 1 contains 3 red and 5 black balls while another Bag 2 contains 4 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it is drawn from bag 2.

(a) 31/62

(b) 16/62

(c) 16/31

(d) 31/32

I had been asked this question in exam.

This interesting question is from Bayes Theorem topic in division Probability of Mathematics – Class 12

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Correct option is (c) 16/31

The explanation is: Let E1 = event of choosing the bag 1, E2 = event of choosing the bag 2.

Let A be event of drawing a red ball.

P(E1) = P(E2) = 1/2.

Also, P(A|E1) = P(drawing a red ball from Bag 1) = 3/8.

And P(A|E2) = P(drawing a red ball from Bag 2) = 4/10.

The probability of drawing a ball from bag 2, being given that it is red is P(E2|A).

By using Bayes’ theorem,

P(E2|A) = P(E2)P(A|E2)/( P(E1)P(A│E1)+P(E2)P(A|E2))

= (1/2 × 4/10) / ((1/2 × 3/8)) + (1/2 × 4/10)) = 16/31.