# A speaks truth in 75% cases and B in 80% of the cases. In what percentage of the cases are they likely to contradict each other, narrating the same incident? (a) 5% (b) 15% (c) 35% (d) 45%

A speaks truth 75%

That is,

P(A) = 3/4

$Pleft&space;(&space;bar{A}&space;right&space;)&space;=&space;frac{1}{4}$

Similarly,

B speaks truth 80%

That is,

P(B) = 4/5

$Pleft&space;(&space;bar{B}&space;right&space;)&space;=&space;frac{1}{5}$

Probability = $P&space;(A)&space;P(bar{B})&space;+&space;P(bar{A})P(B)$

On substituting, we get,

= (3/4) (1/5) + (1/4) (4/5)

= 7/20

= 7/20 x 100%

We get,

= 35%

Hence, the correct answer is option (c)

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