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

A. 5%

B. 15%

C. 35%

D. 45%

Solution:

Let

E_{1}

= event that A speaks the truth
And

E_{2}

= event that B speaks the truth
Then,

\begin{aligned} P (E_{1}) = \frac{75}{100} = \frac{3}{4}, \\ P (E_{2}) = \frac{80}{100} = \frac{4}{5}, \\ P ({\bar{E}}_{1}) = (1 - \frac{3}{4}) = \frac{1}{4}, \\ P ({\bar{E}}_{2}) = (1 - \frac{4}{5}) = \frac{1}{5} \end{aligned}

P (A and B contradict each other)
= P [(A speaks the truth and B tells a lie) or (A tells a lie and B speaks the truth)]

\begin{aligned} = P [(E_{1} \cap {\bar{E}}_{2}) o r ({\bar{E}}_{1} \cap E_{2})] \\ = P [(E_{1} \cap {\bar{E}}_{2}) + ({\bar{E}}_{1} \cap E_{2})] \\ = P (E_{1}) . P ({\bar{E}}_{2}) + P ({\bar{E}}_{1}) . P (E_{2}) \\ = (\frac{3}{4} \times \frac{1}{5}) + (\frac{1}{4} \times \frac{4}{5}) \\ = (\frac{3}{20} + \frac{1}{5}) \\ = \frac{7}{20} \\ = (\frac{7}{20} \times 100) % \\ = 35 % \end{aligned}

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