Question

In shuffling a pack of 52 playing cards, four are accidently dropped; find the chance that the missing cards should be one from each suit.

Answer 1

It is given that from a well-shuffled pack of cards, four cards are missing out.
∴ Total number of elementary events, n(S) = ⁵²C₄
Let E be the event where four cards are missing from each suit.
i.e. n(E)= ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁
Hence, required probability = \[\frac{n\left( E \right)}{n\left( S \right)} = \frac{^{13}{}{C}_1 \times ^{13}{}{C}_1 \times ^{13}{}{C}_1 \times ^{13}{}{C}_1}{^{52}{}{C}_4}\]

\[= \frac{13 \times 13 \times 13 \times 13}{\frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1}}\]
\[ = \frac{13 \times 13 \times 13 \times 13}{13 \times 17 \times 25 \times 49}\]
\[ = \frac{2197}{20825}\]