Question

A pack of cards contains 4 aces, 4 kings, 4 queens and 4 jacks. Two cards are drawn at random. The probability that at least one them is an ace is 

(a) \(\frac{1}{5}\) 

(b) \(\frac{1}{9}\)

(c) \(\frac{3}{16}\)

(d) \(\frac{9}{20}\)

Answer 1

(d) \(\frac{9}{20}\)

Let S be the sample space for drawing 2 cards out of 4 aces, 4 kings, 4 queens and 4 jacks i.e, 16 cards. 

Then n(S) = ¹⁶C₂ 

P(Drawing at least one ace) = 1 – P(Drawing no ace) 

Let E : Event of drawing no aces in the 2 drawn cards 

⇒ n(E) = ¹²C₂(Cards leaving aces = 16 – 4 – 12)

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{^{12}C_2}{^{16}C_2}\) = \(\frac{12\times11}{16\times15}\) = \(\frac{11}{20}\)

∴ P(drawing at least one ace) = 1 - \(\frac{11}{20}\) = \(\frac{9}{20}\) .