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 cards is black, is

$\frac{2}{3}$

Let A₁ be the event that the black card is lost, A₂ be the event that the red card is lost and let E be the event that first 13 cards examined are red.

Then the required probability $=P\left(\frac{A_1}{E}\right)$. We have $P(A_1)=P(A_2)=\frac{1}{2};$ as black and red cards were initially equal in number.

Also $P\left(\frac{E}{A_1}\right)=\frac{^{26}C_{13}}{^{51}C_{13}} $ and $P\left(\frac{E}{A_2}\right)=\frac{^{25}C_{13}}{^{51}C_{13}} $

The required probability $=P\left(\frac{A_1}{E}\right)= \frac{P(E/A_1)P(A_1)}{P(E/A_1)P(A_1)+P(E/A_2)P(A_2)}=\frac{\frac{1}{2}.\frac{^{26}C_{13}}{^{51}C_{13}}}{\frac{1}{2}.\frac{^{26}C_{13}}{^{51}C_{13}}+\frac{1}{2}.\frac{^{25}C_{13}}{^{51}C_{13}}}=\frac{2}{3}$