In a class,50 students play cricket,20 studen | vedclass

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(A) Let $C$ be the set of students who play cricket and $F$ be the set of students who play football.Given:$n(C) = 50$$n(F) = 20$$n(C \cap F) = 10$We

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) Let $C$ be the set of students who play cricket and $F$ be the set of students who play football.Given:$n(C) = 50$$n(F) = 20$$n(C \cap F) = 10$We need to find the number of students who play at least one of these two games,which is $n(C \cup F)$.Using the formula: $n(C \cup F) = n(C) + n(F) - n(C \cap F)$$n(C \cup F) = 50 + 20 - 10 = 60$Alternatively,from the Venn diagram:Students playing only cricket = $50 - 10 = 40$Students playing only football = $20 - 10 = 10$Students playing both = $10$Total students playing at least one game = $40 + 10 + 10 = 60$.

In a class,$50$ students play cricket,$20$ students play football and $10$ play both cricket and football. How many students play at least one of these two games?

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