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Question
In a group of students, there are 3 boys and 4 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.
Solution
The group consists of 3 boys and 4 girls i.e., 7 students.
4 students can be selected from this group in `""^7"C"_4 = (7 xx 6 xx 5 xx 4)/(4 xx 3 xx 2 xx 1)` = 35 ways
∴ n(S) = 35
Let A be the event that 3 boys and 1 girl are selected.
3 boys can be selected in 3C3 ways while a girl can be selected in 4C1ways.
∴ n(A) = `""^3"C"_3 xx ""^4"C"_1` = 4
∴ P(A) = `("n"("A"))/("n"("S")) = 4/35`
Let B be the event that 3 girls and 1 boy are selected.
3 girls can be selected in 4C3 ways and a boy can be selected in 3C1 ways.
∴ n(B) = `""^3"C"_1 xx""^4"C"_3 = ""^3"C"_1 xx ""^4"C"_1` = 3 × 4 = 12
∴ P(B) = `("n"("B"))/("n"("S")) = 12/35`
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ Required probability = P(A ∪ B)
= P(A) + P(B)
= `4/35 + 12/35`
= `16/35`
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