Question

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

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{{}^{25}{C}_{13}}{{}^{15}{C}_{13}}$

Answer 1

The correct option is B $\frac{2}{3}$

Let $E_1$ be the event of missing black card and $E_2$ be the event of missing red card.
Then$P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$
Let $A$ be the event that first $13$ cards observed are red
$P\left(\frac{A}{E_1}\right)=\frac{{}^{26}{C}_{13}}{{}^{51}{C}_{13}}\text{and}P\left(\frac{A}{E_2}\right)=\frac{{}^{25}{C}_{13}}{{}^{51}{C}_{13}}$
$P\left(E_1/A\right)=\left[\frac{P\left(E_1\right).P\left(A/E_1\right)}{P\left(E_1\right).P\left(A/E_1\right)+P\left(E_2\right).P\left(A/E_2\right)}\right]$
$=\frac{\frac{1}{2}\left[\frac{\left({}^{26}{C}_{13}\right)}{\left({}^{51}{C}_{13}\right)}\right]}{\frac{1}{2}\left[\left(\frac{{}^{26}{C}_{13}}{{}^{51}{C}_{13}}\right)\right]+\left[\left(\frac{{}^{25}{C}_{13}}{{}^{51}{C}_{13}}\right)\right]}=\left(\frac{\frac{26}{13}{\times }^{25}{C}_{12}}{\frac{26}{13}{\times }^{25}{C}_{12}{+}^{25}{C}_{12}}\right)=\left(\frac{2}{2+1}\right)=\frac{2}{3}$