The probability that A speaks truth is $4 / 5$, while the probability that $B$ speaks truth is $3 / 4$. Then…

The probability that A speaks truth is $4 / 5$, while the probability that $B$ speaks truth is $3 / 4$. Then the probability that A and B contradict each other, when asked to reveal the fact is
  1. $\frac{3}{20}$
  2. $\frac{1}{5}$
  3. $\frac{7}{20}$
  4. $\frac{4}{5}$

Solution

$A=$ Event that A speaks the truth $B=$ Event that B speaks the truth $\begin{aligned} & P(A)=\frac{4}{5} \Rightarrow P\left(A^c\right)=\frac{1}{5} \\ & P(B)=\frac{3}{4} \Rightarrow P\left(B^c\right)=\frac{1}{4}\end{aligned}$ Now, the required probability $=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}\left(\mathrm{B}^{\mathrm{c}}\right)+\mathrm{P}\left(\mathrm{A}^{\mathrm{c}}\right)$ $\mathrm{P}(\mathrm{B})$ $=\frac{4}{5} \times \frac{1}{4}+\frac{1}{5} \times \frac{3}{4}=\frac{7}{20}$

Asked in: AP EAMCET 2023 (18 May Shift 1)

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