In a class of $60$ students, $30$ opted for $NCC$ , $32$ opted for $NSS$ and $24$ opted for both $NCC$ and $NSS$. If one of these students is selected at random, find the probability that The student has opted neither $NCC$ nor $NSS$.
In a class of $60$ students, $30$ opted for $NCC$ , $32$ opted for $NSS$ and $24$ opted for both $NCC$ and $NSS$. If one of these students is selected at random, find the probability that The student has opted neither $NCC$ nor $NSS$.
Let $A$ be the event in which the selected student has opted for $NCC$ and $B$ be the event in which the selected student has opted for $NSS$.
Total number of students $=60$
Number of students who have opted for $NCC =30$
$\therefore $ $P(A)=\frac{30}{60}=\frac{1}{2}$
Number of students who have opted for $NSS =32$
$\therefore $ $P(B)=\frac{32}{60}=\frac{8}{15}$
Number of students who have opted for both $NCC$ and $NSS = 24$
$\therefore $ $P ( A$ and $B )=\frac{24}{60}=\frac{2}{5}$
$P ($ not $A$ and not $B)$
$= P(A ^{\prime}$ and $B ^{\prime})$
$= P \left( A^{\prime} \cap B ^{\prime}\right)$
$= P ( A \cup B )^{\prime}$ $[( A^{\prime} \cap B )=( A \cup B )^{\prime}$ by De Morgan's law $)]$
$=1- P ( A \cup B )$
$=1- P ( A$ or $B )$
$=1-\frac{19}{30}$
$=\frac{11}{30}$
Thus, the probability that the selected students has neither opted for $NCC$ nor $NSS$ is $\frac{11}{30}$
Similar Questions
True statement $A$ and true statement $B$ are two independent events of an experiment.Let $P\left( A \right) = 0.3$ , $P\left( {A \vee B} \right) = 0.8$ then $P\left( {A \to B} \right)$ is (where $P(X)$ denotes probability that statement $X$ is true statement)
True statement $A$ and true statement $B$ are two independent events of an experiment.Let $P\left( A \right) = 0.3$ , $P\left( {A \vee B} \right) = 0.8$ then $P\left( {A \to B} \right)$ is (where $P(X)$ denotes probability that statement $X$ is true statement)
Given two mutually exclusive events $A$ and $B$ such that $P(A) = 0.45$ and $P(B) = 0.35,$ then $P (A$ or $B ) =$
Given two mutually exclusive events $A$ and $B$ such that $P(A) = 0.45$ and $P(B) = 0.35,$ then $P (A$ or $B ) =$
If $A$ and $B$ are arbitrary events, then
If $A$ and $B$ are arbitrary events, then
An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on the is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered $1, 2, 3,….., 9$ is randomly picked and the number on the card is noted. The probability that the noted number is either $7$ or $8$ is
An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on the is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered $1, 2, 3,….., 9$ is randomly picked and the number on the card is noted. The probability that the noted number is either $7$ or $8$ is
- [JEE MAIN 2019]
An unbiased die is thrown twice. Let the event $A$ be 'odd number on the first throw' and $B$ the event 'odd number on the second throw '. Check the independence of the events $A$ and $B$.
An unbiased die is thrown twice. Let the event $A$ be 'odd number on the first throw' and $B$ the event 'odd number on the second throw '. Check the independence of the events $A$ and $B$.