A man is known to speak truth 8 out of 11 times. He takes out two balls from a bag containing 3 white, 2 black and 5 red balls. What is the probability that he reports both the balls to be white?
A man is known to speak truth 8 out of 11 times. He takes out two balls from a bag containing 3 white, 2 black and 5 red balls. What is the probability that he reports both the balls to be white?
Solution
Assumption: When the man lies, he reports the opposite of what he sees for each ball (white becomes non-white, non-white becomes white).
Step 1: Basic probabilities
Total balls = 3 + 2 + 5 = 10
Ways to draw 2 balls = ¹⁰C₂ = 45
- P(truth) = 8/11
- P(lie) = 3/11
Step 2: Case 1, both balls are white and he tells the truth
Ways to draw 2 white balls = ³C₂ = 3
P(both white) = 3/45 = 1/15
P(Case 1) = 1/15 × 8/11 = 8/165
Step 3: Case 2, neither ball is white and he lies
Non-white balls = 2 + 5 = 7
Ways to draw 2 non-white balls = ⁷C₂ = 21
P(neither white) = 21/45 = 7/15
Since he lies, "both non-white" becomes "both white."
P(Case 2) = 7/15 × 3/11 = 21/165
Step 4: Why the mixed case is ignored
If he draws one white and one non-white ball and lies, he reports one non-white and one white. That is still not "both white," so this case does not count.
Step 5: Add the cases
P(reports both white) = 8/165 + 21/165 = 29/165
Correct option: (3) 29/165
(The answer given by NTA is option 3 - which can only come through the assumption written in the start of the answer).
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