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

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

Similarly,

B speaks truth 80%

That is,

P(B) = 4/5

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

While contradicting the narration

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

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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