Question

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

• A. 0.35
• B. 0.45
• C. 0.15
• D. 0.05

Answer 1

Answer: (A)

0.35

Answer 2

$P\left(A\right) = \frac{75}{100} = \frac{3}{4}, P\left(B\right) = \frac{80}{100}= \frac{4}{5}$ $\Rightarrow P\left(\bar{A} \right) = 1 -P\left(A\right)$ and $P\left(\bar{B}\right) = 1 - P\left(B\right)$ $\Rightarrow P\left(\bar{A} \right)= 1 - \frac{3}{4} = \frac{1}{4}$ and $P\left(\bar{B} \right) = 1 - \frac{4}{5} = \frac{1}{5}$ Now, prob. (contradict to each other) $= P\left(A \bar{B} or \bar{A }B\right)=P\left(A \bar{B} \right) + P\left(\bar{A} B\right)$ $= P\left(A\right)P\left(\bar{B} \right) + P\left(\bar{A} \right) P\left(B\right) = \frac{3}{4} . \frac{1}{5} + \frac{1}{4} . \frac{4}{5}$ $\Rightarrow P\left(A\bar{B} or \bar{A} B\right) = \frac{7}{20}$ $\therefore$%