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Probability

Question
CBSEENMA12033718
Wired Faculty App

A speaks truth in 75% and B in 80% of the cases. In what percentage of cases are they likely to contradict each other in narrating the same incident?

Solution

Let P(A), P(B) be the probability of A and B are speaking the truths. Then
              straight P left parenthesis straight A right parenthesis space equals space 75 over 100 space equals 3 over 4 comma space space space straight P left parenthesis straight B right parenthesis space equals 80 over 100 space equals 4 over 5
therefore space space space space space straight P left parenthesis straight A with bar on top right parenthesis space equals space straight P left parenthesis straight A space tells space straight a space lie right parenthesis space equals space 1 space space minus straight P left parenthesis straight A right parenthesis space equals space 1 minus space 3 over 4 space equals 1 fourth space space space space space space space space space straight P left parenthesis straight B with bar on top right parenthesis space equals space straight P left parenthesis straight B space tells space straight a space lie right parenthesis space equals space 1 minus straight P left parenthesis straight B right parenthesis space equals space 1 minus 4 over 5 space equals 1 fifth
Now, P(A and B contradict) = straight P left parenthesis straight A right parenthesis space straight P left parenthesis straight B with bar on top right parenthesis space plus space straight P left parenthesis straight A with bar on top right parenthesis space straight P left parenthesis straight B right parenthesis
space equals space 3 over 4 cross times 1 fifth plus 4 over 5 cross times 1 fourth space equals space 3 over 20 plus 1 fifth space equals 7 over 20 equals space 35 percent sign

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