Question

Five cards are drawn from a pack of 52 cards. What is the chance that these 5 will contain:
(i) just one ace

Answer 1

Let S denote the sample space.
Then n(S) = 52
Thus, five cards can be drawn in ⁵²C₅ ways.
∴ Total number of elementary events = ⁵²C₅

(i)
Let E₁ = event of getting just one ace
i.e. E₁ = ⁴C₁ × ⁴⁸C₄ 
∴ Total number of favourable events = ⁴C₁ × ⁴⁸C₄ 
       Hence, required probability = \[\frac{^{4}{}{C}_1 \times ^{48}{}{C}_4}{^{52}{}{C}_5}\]

\[= \frac{4 \times 2 \times 47 \times 46 \times 45}{52 \times 51 \times 10 \times 49 \times 2}\]

\[= \frac{47 \times 46 \times 9}{13 \times 51 \times 2 \times 49}\]

\[= \frac{47 \times 23 \times 3}{13 \times 17 \times 49} = \frac{3243}{10829}\]