$A$ speaks truth in $75 \%$ of the cases and $B$ in $80 \%$ of the cases. Then, the probability that their…

$A$ speaks truth in $75 \%$ of the cases and $B$ in $80 \%$ of the cases. Then, the probability that their statements about an incident do not match, is
  1. $\frac{7}{20}$
  2. $\frac{3}{20}$
  3. $\frac{2}{7}$
  4. $\frac{5}{7}$

Solution

Let Event A: A speaks the truth Event B: B speaks the truth $\begin{aligned} & P(A)=75 \%=\frac{75}{100}=\frac{3}{4} \\ & \text { and } P(B)=80 \%=\frac{80}{100}=\frac{4}{5} \end{aligned}$ $\therefore \quad$ Required probability $\begin{aligned} & P(A \bar{B})+P(\bar{A} B)=P(A) \times P(\bar{B})+P(\bar{A}) \times P(B) \\ & =P(A) \times[1-P(B)]+P(A) \times P(B) \\ & =\frac{3}{4} \times\left(1-\frac{4}{5}\right)+\left(1-\frac{3}{4}\right) \times \frac{4}{5} \\ & =\frac{3}{4} \times \frac{1}{5}+\frac{1}{4} \times \frac{4}{5}=\frac{7}{20} \end{aligned}$

Asked in: AP EAMCET 2016

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