Question

From a pack of 52 cards, 4 are drawn one by one without replacement. Find the probability that all are aces.

Answer 1

Consider the given events.
A = An ace in the first draw
B = An ace in the second draw
C = An ace in the third draw
D = An ace in the fourth draw

$$\begin{gathered} \mathrm{Now}, \\ P\left(A\right)=\frac{4}{52}=\frac{1}{13} \\ P\left(B/A\right)=\frac{3}{51}=\frac{1}{17} \\ P\left(C/A\cap B\right)=\frac{2}{50}=\frac{1}{25} \\ P\left(D/A\cap B\cap C\right)=\frac{1}{49} \\ \therefore \mathrm{Required}\mathrm{probability}=P\left(A\cap B\cap C\cap D\right)=P\left(A\right)\times P\left(B/A\right)\times P\left(C/A\cap B\right)\times P\left(D/A\cap B\cap C\right) \\ =\frac{1}{13}\times \frac{1}{17}\times \frac{1}{25}\times \frac{1}{49} \\ =\frac{1}{270725} \end{gathered}$$