Question

From a deck of 52 cards,4 Cards are drawn at random without replacing
What is the Probability of getting 3 consecutive aces

Answer 1

$$\begin{gathered} \mathrm{Hi}, \\ \mathrm{The}\mathrm{required}\mathrm{can}\mathrm{be} \\ \mathrm{ACE},\mathrm{ACE},\mathrm{ACE},\mathrm{non}\mathrm{ACE} \\ \mathrm{non}\mathrm{ACE},\mathrm{ACE},\mathrm{ACE},\mathrm{ACE} \\ \mathrm{IN}\mathrm{first}\mathrm{case}, \\ \mathrm{Drawing}\mathrm{first}\mathrm{ace}\mathrm{will}\mathrm{give}\mathrm{probability}=\frac{4}{52} \\ \mathrm{Drawing}\mathrm{second}\mathrm{ace}\mathrm{will}\mathrm{give}\mathrm{probability}=\frac{3}{51} \\ \mathrm{Drawing}\mathrm{third}\mathrm{ace}\mathrm{will}\mathrm{give}\mathrm{probability}=\frac{2}{50} \\ \mathrm{Drawing}\mathrm{fourth}\mathrm{non}\mathrm{ACE}=\frac{1}{49} \\ \mathrm{SO}\mathrm{required}\mathrm{probability}=\frac{4}{52}\times \frac{3}{51}\times \frac{2}{50}\times \frac{1}{49}=\frac{1}{270725} \\ \mathrm{Now}\mathrm{in}\mathrm{second}\mathrm{case} \\ =\frac{48}{52}\times \frac{4}{51}\times \frac{3}{50}\times \frac{2}{49}=\frac{48}{270725} \\ \mathrm{So}\mathrm{required}\mathrm{probability}=\frac{48}{270725}+\frac{1}{270725}=\frac{49}{270725} \end{gathered}$$