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 = $\frac{n\left(E\right)}{n\left(S\right)}=\frac{{}^{13}C_1\times {}^{13}C_1\times {}^{13}C_1\times {}^{13}C_1}{{}^{52}C_4}$

$$\begin{gathered} =\frac{13\times 13\times 13\times 13}{{\displaystyle \frac{52\times 51\times 50\times 49}{4\times 3\times 2\times 1}}} \\ =\frac{13\times 13\times 13\times 13}{13\times 17\times 25\times 49} \\ =\frac{2197}{20825} \end{gathered}$$