# 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

$$\begin{array}{l}P(\bar{A})=\frac{1}{4}\end{array}$$

Similarly,

B speaks truth 80%

That is,

P(B) = 4/5

$$\begin{array}{l}P(\bar{B})=\frac{1}{5}\end{array}$$

$$\begin{array}{l}Probability = P(A)P(\bar{B})+P(\bar{A})P(B)\end{array}$$

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)