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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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.

Similar Questions

One card is drawn at random from a pack of $52$ playing cards. The probability that it is an ace,a black king,or the queen of hearts is:

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three persons apply for the same house is

If the events $A$ and $B$ are independent,and $P(A^{\prime}) = \frac{2}{3}$ and $P(B^{\prime}) = \frac{2}{7}$,then $P(A \cap B)$ is equal to:

One card is drawn from a well-shuffled deck of $52$ cards. If each outcome is equally likely,calculate the probability that the card will be a diamond.

The probability that two randomly selected subsets of the set $\{1, 2, 3, 4, 5\}$ have exactly two elements in their intersection is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module