Question

From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected.

(a) \(\frac{1}{5}\)

(b) \(\frac{7}{10}\)

(c) \(\frac{2}{5}\) 

(d) \(\frac{5}{6}\)

Answer 1

(b) \(\frac{7}{10}\)

Total number of ways of selecting 2 persons at random out of 5 persons = ⁵C₂ 

∴ n(S) = ⁵C₂  = \(\frac{|\underline5}{|\underline3|\underline2}\) = \(\frac{5\times4}{2\times1}\) = 10

Let A : Event of selecting at least one woman.

⇒ A = Selecting 1 woman + 1 man or selecting 2 women 

⇒ n(A) = ²C₁ × ³C₁ + ²C₂ = 2 × 3 + 1 = 7 

∴ Required probability = \(\frac{n(A)}{n(S)}\) = \(\frac{7}{10}\).