A person goes to office by car, scooter, bus, | vedclass

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

Question

(A) Let $A, B, C, D$ be the events that the person goes to the office by car, scooter, bus, and train respectively. Then $P(A) = 1/7, P(B) = 3/7, P(C)

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $A, B, C, D$ be the events that the person goes to the office by car, scooter, bus, and train respectively. Then $P(A) = 1/7, P(B) = 3/7, P(C) = 2/7, P(D) = 1/7$.Let $L$ be the event that he reaches late and $E$ be the event that he reaches in time. Then $P(E|A) = 1 - P(L|A) = 1 - 2/9 = 7/9$. Similarly, $P(E|B) = 1 - 1/9 = 8/9, P(E|C) = 1 - 4/9 = 5/9, P(E|D) = 1 - 1/9 = 8/9$.We need to find $P(A|E)$. By Bayes' Theorem:$P(A|E) = \frac{P(A)P(E|A)}{P(A)P(E|A) + P(B)P(E|B) + P(C)P(E|C) + P(D)P(E|D)}$$P(A|E) = \frac{(1/7)(7/9)}{(1/7)(7/9) + (3/7)(8/9) + (2/7)(5/9) + (1/7)(8/9)}$$P(A|E) = \frac{7/63}{7/63 + 24/63 + 10/63 + 8/63} = \frac{7}{7 + 24 + 10 + 8} = \frac{7}{49} = 1/7$.

Similar Questions

Bag $A$ contains $2$ white and $3$ red balls. Bag $B$ contains $4$ white and $5$ red balls. One bag is chosen at random and a ball is drawn from it,which is found to be red. Find the probability that it was drawn from bag $B$.

An urn contains $6$ white and $9$ black balls. Two successive draws of $4$ balls are made without replacement. The probability,that the first draw gives all white balls and the second draw gives all black balls,is:

$A$ bag $P$ contains $5$ white marbles and $3$ black marbles. Four marbles are drawn at random from $P$ and are put in an empty bag $Q$. If a marble drawn at random from $Q$ is found to be black,then the probability that all the three black marbles in $P$ were transferred to the bag $Q$ is:

$A$ bag $A$ contains $2$ white and $3$ red balls and bag $B$ contains $4$ white and $5$ red balls. One ball is drawn at random from a randomly chosen bag and is found to be red. The probability that it was drawn from bag $B$ is:

Events $E_{1}$ and $E_{2}$ form a partition of the sample space $S$. $A$ is any event such that $P(E_{1}) = P(E_{2}) = \frac{1}{2}$,$P(E_{2} | A) = \frac{1}{2}$,and $P(A | E_{2}) = \frac{2}{3}$. Then $P(E_{1} | A)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module