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

The probability of obtaining an even prime number on each die,when a pair of dice is rolled,is . . . . . . .

Three coins are tossed. Describe two events which are mutually exclusive but not exhaustive.

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

Let $A$ and $B$ be events for which $P(A) = x$,$P(B) = y$,and $P(A \cap B) = z$. Then $P(\bar{A} \cap B)$ equals:

Out of $100$ students,two sections of $40$ and $60$ are formed. If you and your friend are among the $100$ students,what is the probability that you both enter different sections?

Vedclass Test Series

Exam Paper Generator

Online Exam Module