One card is missing in a pack of 52 playing cards. If two cards are drawn randomly from the remaining cards…
One card is missing in a pack of 52 playing cards. If two cards are drawn randomly from the remaining cards at a time and are found to be spades, then the probability that the missing card is not a spade is
$\frac{3}{50}$
$\frac{39}{50}$
$\frac{39}{52}$
$\frac{38}{52}$
Solution
Let $\mathrm{E}_1=$ Event of choosing a spade card
$\mathrm{E}_2=$ Event of choosing a card which is not spade
Let A denote the lost card.
$\therefore \quad \mathrm{P}\left(\mathrm{E}_1\right)=\frac{13}{52}=\frac{1}{4}$
$\begin{aligned} & P\left(E_2\right)=\frac{39}{52}=\frac{3}{4} \\ & P\left(\frac{A}{E_1}\right)=\frac{{ }^{12} C_2}{{ }^{51} C_2}=\frac{22}{425} \\ & P\left(\frac{A}{E_2}\right)=\frac{{ }^{13} C_2}{{ }^{51} C_2}=\frac{26}{425}\end{aligned}$
By using Baye's theorem,
$P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right) P\left(\frac{A}{E_2}\right)}{P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)}$
$\begin{aligned} & =\frac{\frac{3}{4} \times \frac{26}{425}}{\frac{1}{4} \times \frac{22}{425}+\frac{3}{4} \times \frac{26}{425}} \\ & =\frac{3 \times 26}{22+3 \times 26}=\frac{39}{50}\end{aligned}$