A boy gets a chance of 55% to win 1st round of a game and a girl gets a chance of 60% to win 2nd round of the game. In what % of cases are they likely to contradict each other, narrating the same inci

manpreet Best Answer 1 year ago

49%

Let A be the EVENT that a boy wins 1ST ROUND
Let B be the event that a girl wins 2nd round.
Then, A' = Event that the boy losses 1st round and B' = event that the girl losses 2nd round.
Therefore, P(A) = 55/100 = 11/20, P(B) = 60/100 = 12/20.
P(A') = 1- (11/20) = 9/20 and P(B') = 1- (12/20) = 8/20
First, we have to find the probability that they contradict each other;
That is, P( A And B contradicts each other)
= P[(boy WIN in 1st round And girl losses in 2nd round) (Or) (boy losses in 1st round And girl wins in 2nd round)
= P[(A And B') Or (A' And B)] = P[(A And B')] + p[(A' And B)]
= P(A) x P(B') + P(A') x P(B)
= 11/20 x 8/20 + 9/20 x 12/20
= 88/400 + 108/400
= 196/400
We have to find the %.
Required % = (196/400)x100 = 49%.