In a class of 60 students,30 opted for NCC,32 | vedclass

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

Question

(A) Let $A$ be the event that the selected student opted for $NCC$ and $B$ be the event that the selected student opted for $NSS$.Total number of stud

Vedclass · Accepted Answer

$\frac{19}{30}$

Vedclass · Answer

(A) Let $A$ be the event that the selected student opted for $NCC$ and $B$ be the event that the selected student opted for $NSS$.Total number of students $n(S) = 60$.Number of students who opted for $NCC$,$n(A) = 30$.Number of students who opted for $NSS$,$n(B) = 32$.Number of students who opted for both $NCC$ and $NSS$,$n(A \cap B) = 24$.We need to find the probability that the student opted for $NCC$ or $NSS$,which is $P(A \cup B)$.Using the formula $n(A \cup B) = n(A) + n(B) - n(A \cap B)$:$n(A \cup B) = 30 + 32 - 24 = 38$.Therefore,$P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{38}{60} = \frac{19}{30}$.

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 opted for $NCC$ or $NSS$.

Similar Questions

In a class of $30$ pupils,$12$ take needle work,$16$ take physics,and $18$ take history. If all the $30$ students take at least one subject and no one takes all three,then the number of pupils taking exactly $2$ subjects is:

In a class of $125$ students,$70$ passed in Mathematics,$55$ in Statistics,and $30$ in both. The probability that a student selected at random from the class has passed in only one subject is

Let $S$ be the set of all ordered pairs $(x, y)$ of positive integers satisfying the condition $x^2 - y^2 = 12345678$. Then,

If $n(A) = 3$,$n(B) = 6$ and $A \subseteq B$. Then the number of elements in $A \cup B$ is equal to

Let $X$ be the set consisting of the first $2018$ terms of the arithmetic progression $1, 6, 11, \dots$ and $Y$ be the set consisting of the first $2018$ terms of the arithmetic progression $9, 16, 23, \dots$. Then,the number of elements in the set $X \cup Y$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module