In a school,there are 21 students in the cric | vedclass

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

Question

(A) Let $C$,$H$,and $F$ represent the sets of students in the cricket,hockey,and football teams,respectively.Given:$n(C) = 21$,$n(H) = 26$,$n(F) = 29$

Vedclass · Accepted Answer

$43$

Vedclass · Answer

(A) Let $C$,$H$,and $F$ represent the sets of students in the cricket,hockey,and football teams,respectively.Given:$n(C) = 21$,$n(H) = 26$,$n(F) = 29$$n(H \cap C) = 14$,$n(H \cap F) = 15$,$n(F \cap C) = 12$$n(H \cap C \cap F) = 8$We need to find the total number of students,which is $n(C \cup H \cup F)$.Using the Principle of Inclusion-Exclusion:$n(C \cup H \cup F) = n(C) + n(H) + n(F) - [n(H \cap C) + n(H \cap F) + n(F \cap C)] + n(H \cap C \cap F)$$n(C \cup H \cup F) = 21 + 26 + 29 - [14 + 15 + 12] + 8$$n(C \cup H \cup F) = 76 - 41 + 8$$n(C \cup H \cup F) = 35 + 8 = 43$Thus,the total number of students is $43$.

Similar Questions

$2n(A \setminus B) = n(B \setminus A)$ and $5n(A \cap B) = n(A) + 3n(B)$,where $P \setminus Q = P \cap Q^C$. If $n(A \cup B) \leq 10$,then the value of $\frac{n(A) \cdot n(B) \cdot n(A \cap B)}{8}$ is:

The probability that at least one of the events $E_1$ and $E_2$ occurs is $0.6$. If the probability of the simultaneous occurrence of $E_1$ and $E_2$ is $0.2$,then $P(E_1') + P(E_2') = $

$A$ survey shows that $63\%$ of the Americans like cheese whereas $76\%$ like apples. If $x\%$ of the Americans like both cheese and apples,then

If $A$ and $B$ are events of a random experiment with $P(A) = 0.5$,$P(B) = 0.4$ and $P(A \cap B) = 0.3$,then the probability that neither $A$ nor $B$ occurs is

In a certain town, $60 \%$ of the families own a car, $30 \%$ own a house and $20 \%$ own both a car and a house. If a family is randomly chosen, what is the probability that this family owns a car or a house but not both?

Vedclass Test Series

Exam Paper Generator

Online Exam Module