The probability that Ajay will not appear in | vedclass

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

Question

(B) Let $A$ be the event that Ajay appears in the exam and $V$ be the event that Vijay appears in the exam.Given: $P(\overline{A}) = \frac{2}{7}$,so $

Vedclass · Accepted Answer

$\frac{18}{35}$

Vedclass · Answer

(B) Let $A$ be the event that Ajay appears in the exam and $V$ be the event that Vijay appears in the exam.Given: $P(\overline{A}) = \frac{2}{7}$,so $P(A) = 1 - P(\overline{A}) = 1 - \frac{2}{7} = \frac{5}{7}$.Given: $P(A \cap V) = \frac{1}{5}$.We need to find the probability that Ajay appears and Vijay does not appear,which is $P(A \cap \overline{V})$.Using the property of sets,$P(A) = P(A \cap V) + P(A \cap \overline{V})$.Substituting the values: $\frac{5}{7} = \frac{1}{5} + P(A \cap \overline{V})$.$P(A \cap \overline{V}) = \frac{5}{7} - \frac{1}{5} = \frac{25 - 7}{35} = \frac{18}{35}$.

The probability that Ajay will not appear in the $JEE$ exam is $p = \frac{2}{7}$,while the probability that both Ajay and Vijay will appear in the exam is $q = \frac{1}{5}$. Then the probability that Ajay will appear in the exam and Vijay will not appear is:

Similar Questions

$A$ bag contains $5$ black balls,$4$ white balls and $3$ red balls. If a ball is selected at random,the probability that it is a black or a red ball,is

Two dice are thrown simultaneously. The probability of getting the sum $2$,$8$,or $12$ is

One card is drawn from each of two ordinary packs of $52$ cards. The probability that at least one of them is an ace of hearts is:

When two dice are thrown,the probability of getting an ordered pair $(x, y)$ such that $x^2+y^2 \leq 25$,where $x$ and $y$ are the numbers that show up on the two dice,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module