(C) Let $n(M)$ be the number of members proficient in Mathematics and $n(S)$ be the number of members proficient in Statistics.Given $n(M \cup S) = 25

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(C) Let $n(M)$ be the number of members proficient in Mathematics and $n(S)$ be the number of members proficient in Statistics.Given $n(M \cup S) = 25$,$n(M) = 19$,and $n(S) = 16$.Using the formula $n(M \cup S) = n(M) + n(S) - n(M \cap S)$:$25 = 19 + 16 - n(M \cap S)$$25 = 35 - n(M \cap S)$$n(M \cap S) = 35 - 25 = 10$.The probability that a person selected at random is proficient in both is $P(M \cap S) = \frac{n(M \cap S)}{n(M \cup S)} = \frac{10}{25} = \frac{2}{5}$.

Class 11 Mathematics Set Theory Word Problem - Set Theory

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In a committee of $25$ members,each member is proficient either in Mathematics or in Statistics or in both. If $19$ of them are proficient in Mathematics and $16$ of them are proficient in Statistics,then the probability that a person selected at random from the committee is proficient in both is

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A
$\frac{1}{5}$
B
$\frac{3}{5}$
C
$\frac{2}{5}$
D
$\frac{1}{2}$

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In a committee of $25$ members,each member is proficient either in Mathematics or in Statistics or in both. If $19$ of them are proficient in Mathematics and $16$ of them are proficient in Statistics,then the probability that a person selected at random from the committee is proficient in both is

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