A speaks the truth 2 out of 3 times and B, 4 out of 5 times. They agree in the assertion that from a bag containing 6 balls of different colours a black ball has been drawn. Find the probability that the statements are true.
Answer 129.
The ball drawn may be black or non-black (any of the remaining 5 different colours) and both may say that it is black.
Let C = event that the ball drawn is black and both agreeing that it is black
P(C) = (1/6)*(2/3)*(4/5) = 4/45.This is the case in which they are both agreeing and speaking the truth.
Let D = event that the ball drawn is non-black (any of the remaining 5 different colours) and both agreeing that it is black but telling a lie.
Both can tell a lie in 5 different ways. To explain this, suppose the non-black ball drawn is blue, they may say it is black, white, red, yellow, green (all lies) of which probability of telling black is (1/5) on which they agree. A may tell a lie that it is black in (1/5)*(1/3) =1/15 ways and B can tell it in (1/5)*(1/5) =1/25 ways. Also there is (5/6) probability of picking up a non-blackball. Thus, probability that a non-black ball is drawn and both agree that it is black thus asserting that it is black but telling a lie is
P(D) = (5/6)*(1/15)*(1/25) = 1/(450).
Now, we have found the probability of two events in which both are asserting that the ball is black, but when event C occurs the assertion is with speaking the truth and when event D occurs, assertion is by telling a lie.
=> the probability of their asserting that the ball is black and speaking the truth
= P(C) /[P(C) + P(D)]
= (4/45) / [(4/45) + (1/450)]
= 40/41.
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