A speaks truth in 20 % of the cases and B in | vedclass

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(B) Let $E_1$ be the event that $A$ speaks the truth. $P(E_1) = \frac{20}{100} = \frac{1}{5}$. Then,$P(\overline{E_1}) = 1 - \frac{1}{5} = \frac{4}{5}

Vedclass · Accepted Answer

$\frac{17}{25}$

Vedclass · Answer

(B) Let $E_1$ be the event that $A$ speaks the truth. $P(E_1) = \frac{20}{100} = \frac{1}{5}$. Then,$P(\overline{E_1}) = 1 - \frac{1}{5} = \frac{4}{5}$. Let $E_2$ be the event that $B$ speaks the truth. $P(E_2) = \frac{80}{100} = \frac{4}{5}$. Then,$P(\overline{E_2}) = 1 - \frac{4}{5} = \frac{1}{5}$. Their statements do not match if ($A$ speaks the truth and $B$ lies) $OR$ ($A$ lies and $B$ speaks the truth). Required Probability $= P(E_1) \cdot P(\overline{E_2}) + P(\overline{E_1}) \cdot P(E_2)$. $= (\frac{1}{5} 	imes \frac{1}{5}) + (\frac{4}{5} 	imes \frac{4}{5})$. $= \frac{1}{25} + \frac{16}{25} = \frac{17}{25}$.

$A$ speaks truth in $20 \%$ of the cases and $B$ in $80 \%$ of the cases. Find the probability that their statements about an incident do not match.

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