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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Use the principle of inclusion-exclusion for two sets: \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \). Identify the sets (Mathematics and Statistics), their sizes, and the size of their union (total committee members). Solve for the number of members in the intersection (proficient in both). The probability is then the ratio of the number of members proficient in both to the total number of members.
Let \( M \) be the set of members proficient in Mathematics, and \( S \) be the set of members proficient in Statistics.
We are given:
Total number of members in the committee, \( n(U) = 25 \)
Number of members proficient in Mathematics, \( n(M) = 19 \)
Number of members proficient in Statistics, \( n(S) = 16 \)
Each member is proficient in either Mathematics or Statistics or both, which means \( n(M \cup S) = 25 \).
We use the formula for the union of two sets:
$$ n(M \cup S) = n(M) + n(S) - n(M \cap S) $$
Substituting the given values:
$$ 25 = 19 + 16 - n(M \cap S) $$
$$ 25 = 35 - n(M \cap S) $$
$$ n(M \cap S) = 35 - 25 = 10 $$
So, the number of members proficient in both Mathematics and Statistics is 10.
The probability that a person selected at random from the committee is proficient in both is given by:
$$ P(\text{proficient in both}) = \frac{\text{Number of members proficient in both}}{\text{Total number of members}} = \frac{n(M \cap S)}{n(U)} $$
$$ P(\text{proficient in both}) = \frac{10}{25} = \frac{2}{5} $$
