Question

A pack of playing cards was found to contain only $51$ cards.If the first $13$ cards which are examined are all red,what is the probability that the missing card is black?

Answer 1

Let $E_1$ be the event that the missing card is black and $E_1$ be the event that the missing card is red.

Let $A$ be the event the first $13$ cards which are examined are all red.

$\therefore P\left(E_1\right)=\frac{1}{2}$ and $P\left(E_2\right)=\frac{1}{2}$

$P\left(A/E_1\right)=$Probability of selecting $13$ red cards, when the missing card is black$=\frac{{}^{26}{C}_{13}}{{}^{51}{C}_{13}}$

$P\left(A/E_2\right)=$Probability of selecting $13$ red cards, when the missing card is red$=\frac{{}^{25}{C}_{13}}{{}^{51}{C}_{13}}$

$\therefore P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)}$

$=\frac{\frac{1}{2}\times \frac{{}^{26}{C}_{13}}{{}^{51}{C}_{13}}}{\frac{1}{2}\times \frac{{}^{26}{C}_{13}}{{}^{51}{C}_{13}}+\frac{1}{2}\times \frac{{}^{25}{C}_{13}}{{}^{51}{C}_{13}}}$

$=\frac{{}^{26}{C}_{13}}{{}^{26}{C}_{13}{+}^{25}{C}_{13}}$

$=\frac{\frac{26!}{13!13!}}{\frac{26!}{13!13!}+\frac{25!}{13!12!}}$

$=\frac{\frac{26\times 25!}{12\times 12!\times 13!}}{\frac{26!\times 25!}{13\times 13!\times 12!}+\frac{25!}{12!\times 13!}}$

$=\frac{2}{2+1}=\frac{2}{3}$