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X speaks truth in 60% and Y in 50% of...

`X` speaks truth in 60% and `Y` in 50% of the cases. Find the probability that they contradict each other narrating the same incident.

Text Solution

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Let event X be that X speaks the truth and event Y be that Y speaks the truth.
According to the question,
`P(X)=0.6and P(Y)=0.5`
`thereforeP(X')=0.4and P(Y')=0.5`
Let event C be "X and Y contradict each other narrating the same incident".
Clearly, event C occurs when one of X and Y speaks the truth and other does not.
`thereforeP(C)=P(XnnY')+P(X'nnY)`
`=P(X)P(Y')+P(X')P(Y)`
`" "("since X and Y are independent")`
`=0.6xx0.5+0.4xx0.5.=0.5`
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