Question

4 couples (husband-wife) attend a party. Two persons are randomly selected from these 8 persons. Find the probability that the selected persons are

(1) husband and wife,
(2) one man and one woman,
(3) one man and one woman who are not husband and wife.

Answer 1

4 couples (husband-wife) attend a party

∴ No. of persons attending a party = 4 × 2 = 8.
Now, the number of primary outcomes of the sample space for the random experiment of selecting 2 persons randomly from 8 persons

n = ⁸C₂ = \(\frac{8×7}{2×1}\) = 28

(1) A = Event that the selected two persons are husband and wife.
4 couples are there.

∴ Favourable outcomes for the event A
is m = ⁴C₁ = 4.

Hence P(A) = \(\frac{m}{n} = \frac{4}{28} = \frac{1}{7}\)

(2) B = Event that in the selected two persons there is one man and one woman. In 8 persons, there are 4 men and 4 women,
Favourable outcomes for the event B is  

m = ⁴C₁ × ⁴C₁ = 4 × 4 = 16

Hence, P(A) =\(\frac{ m}{n} = \frac{16}{28} = \frac{4}{7}\)

(3) C = Event that in the selected two persons there is one man and one woman who are not husband and wife.

Let 4 couples be denoted by M₁F₁, M₂F₂, M₃F₃, M₄F₄ In selected two persons if a man is selected from a couple and a woman is selected from the remaining couples, then they are not husband and wife.

∴ C = {M₁F₂, M₁F₃, M₁F₄, M₂F₁, M₂F₃, M₂F₄, M₃F₁, M₃F₂, M₃F₄, M₄F₁, M₄F₂, M₄F₃}

∴ Favourable outcomes for the event C is m = 12

Hence, P(C) =\(\frac{ m}{n} = \frac{12}{28} = \frac{3}{7}\)

OR

C = B – A and A ⊂ B

∴ P(C) = P(B) – P(A)

= \(\frac{4}{7} – \frac{1}{7} = \frac{3}{7}\)