The probabilities that A and B speak the trut | vedclass

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

Question

(D) Let $P(A)$ be the probability that $A$ speaks the truth and $P(B)$ be the probability that $B$ speaks the truth. Given,$P(A) = \frac{4}{5}$ and $P

Vedclass · Accepted Answer

$\frac{7}{20}$

Vedclass · Answer

(D) Let $P(A)$ be the probability that $A$ speaks the truth and $P(B)$ be the probability that $B$ speaks the truth. Given,$P(A) = \frac{4}{5}$ and $P(B) = \frac{3}{4}$. The probability that $A$ does not speak the truth is $P(\overline{A}) = 1 - P(A) = 1 - \frac{4}{5} = \frac{1}{5}$. The probability that $B$ does not speak the truth is $P(\overline{B}) = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4}$. They contradict each other if one speaks the truth and the other lies. Thus,the probability of contradiction is $P(A \cap \overline{B}) + P(\overline{A} \cap B) = P(A) \cdot P(\overline{B}) + P(\overline{A}) \cdot P(B)$. Substituting the values: $(\frac{4}{5} 	imes \frac{1}{4}) + (\frac{1}{5} 	imes \frac{3}{4}) = \frac{4}{20} + \frac{3}{20} = \frac{7}{20}$.

The probabilities that $A$ and $B$ speak the truth are $\frac{4}{5}$ and $\frac{3}{4}$ respectively. The probability that they contradict each other when asked to speak on a fact is

Similar Questions

Three critics review a book. For the three critics,the odds in favour of the book are $(5: 2)$,$(4: 3)$ and $(3: 4)$ respectively. The probability that the majority is in favour of the book is

In a certain population,$10\%$ of the people are rich,$5\%$ are famous,and $3\%$ are both rich and famous. The probability that a person picked at random from the population is either famous or rich but not both,is equal to

The three ships namely $A$,$B$ and $C$ sail from India to Africa. If the odds in favour of the ships reaching safely are $2:5$,$3:7$ and $6:11$ respectively,then the probability of all of them arriving safely is

One card is drawn randomly from a pack of $52$ cards. The probability that it is a king or a spade is:

The probability that a leap year selected at random contains either $53$ Sundays or $53$ Mondays is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module