Question

From a pack of 52 playing cards all the face cards are removed. from the remaining pack cards are dealt one by one until an ace appears. What is the probability that the first ace appears in one of the 20 cards?

Answer 1

Dear Student,

Total Cards = 52
Total No. of Face Cards = 12

Thus, No. of Cards Left = 52-12 = 40

$$\begin{gathered} \mathrm{Probability}\mathrm{of}\mathrm{a}\mathrm{getting}\mathrm{an}\mathrm{ACE}\hspace{0.17em}=\frac{4}{40}=\frac{1}{10} \\ \mathrm{Probability}\mathrm{of}\mathrm{not}\mathrm{getting}\mathrm{an}\mathrm{ACE}\hspace{0.17em}=\frac{9}{10} \\ \mathrm{Thus}, \\ \mathrm{Probability}\mathrm{of}\mathrm{a}\mathrm{getting}\mathrm{an}\mathrm{ACE}\mathrm{in}\mathrm{the}1\mathrm{st}\mathrm{Draw}\hspace{0.17em}=\frac{1}{10} \\ \mathrm{For}2\mathrm{nd}\mathrm{Draw},\mathrm{we}\mathrm{must}\mathrm{get}\mathrm{a}\mathrm{non}-\mathrm{ACE}\mathrm{card}\mathrm{on}\mathrm{the}1\mathrm{st}\mathrm{draw} \\ \mathrm{Probability}\mathrm{of}\mathrm{a}\mathrm{getting}\mathrm{an}\mathrm{ACE}\mathrm{in}\mathrm{the}2\mathrm{nd}\mathrm{Draw}\hspace{0.17em}=\left(\frac{9}{10}\right)\left(\frac{1}{10}\right) \\ \mathrm{Probability}\mathrm{of}\mathrm{a}\mathrm{getting}\mathrm{an}\mathrm{ACE}\mathrm{in}\mathrm{the}3\mathrm{rd}\mathrm{Draw}\hspace{0.17em}={\left(\frac{9}{10}\right)}^2\left(\frac{1}{10}\right) \\ \mathrm{Probability}\mathrm{of}\mathrm{a}\mathrm{getting}\mathrm{an}\mathrm{ACE}\mathrm{in}\mathrm{the}20\mathrm{th}\mathrm{Draw}\hspace{0.17em}={\left(\frac{9}{10}\right)}^{19}\left(\frac{1}{10}\right) \\ \mathrm{Thus},\mathrm{Probability}\mathrm{of}\mathrm{a}\mathrm{getting}\mathrm{an}\mathrm{ACE}\mathrm{in}\mathrm{either}\mathrm{of}20\mathrm{Draws}\hspace{0.17em}=\mathrm{P} \\ \mathrm{P}=\frac{1}{10}+\left(\frac{9}{10}\right)\left(\frac{1}{10}\right)+{\left(\frac{9}{10}\right)}^2\left(\frac{1}{10}\right)+...+{\left(\frac{9}{10}\right)}^{19}\left(\frac{1}{10}\right) \\ \mathrm{P}\cong 0.878 \end{gathered}$$