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 ^{3}C_{3 }ways while a girl can be selected in ^{4}C_{1}ways.

∴ 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^{ 4}C_{3} ways and a boy can be selected in ^{3}C_{1} 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`