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

A
$\frac{1}{3}$
B
$\frac{2}{3}$
C
$\frac{1}{2}$
D
$\frac{^{25} C_{13}}{^{51} C_{13}}$

Solution:

Let $A_{1}$ be the event that the black card is lost, $A_{2}$

the event that red card is lost and let $E$ be the event that first 13

cards examined are red. Then,

Required probability $= P (A_{1} / E)$

We have,

$P (A_{1}) = P (A_{2}) = 1 / 2$ , as black and red cards were initially equal in number.

Also, $P (E / A_{1}) = \frac{^{26} C_{13}}{^{51} C_{13}}$ and $P (E / A_{2}) = \frac{^{25} C_{13}}{^{51} C_{13}}$

$∴$ Required probability $= P (A_{1} / E)$

$\begin{array}{l} = \frac{P (E / A_{1}) P (A_{1})}{P (E / A_{1}) P (A_{1}) + P (E / A_{2}) P (A_{2})} \\ \Rightarrow P (A_{1} / E) = \frac{\frac{1}{2} \times \frac{^{26} C_{13}}{^{51} C_{13}}}{\frac{1}{2} \times \frac{^{26} C_{13}}{^{51} C_{13}} + \frac{1}{2} \times \frac{^{25} C_{13}}{^{51} C_{13}}} = \frac{^{26} C_{13}}{^{26} C_{13} +^{25} C_{13}} \\ \Rightarrow P (A_{1} / E) = \frac{\frac{26}{13} (^{25} C_{12})}{\frac{26}{13} (^{25} C_{12}) +^{25} C_{13}} = \frac{2}{2 + 1} = \frac{2}{3} \end{array}$

Solution:

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