Question

A pack of cards is counted with face downwards. It is found that one card is missing, One card is drawn and is found to be red. Find the probability that the missing card is red.

Answer 1

Let $A$ be the event of drawing a red card when one card is drawn out of $51$ cards (excluding missing card.) Let $A_1$ be the event that the missing card is red and $A_2$ be the event that the missing card is black.
Now by Bayes's theorem, required probability,
$P\left(A_1/A\right)=\frac{P\left(A_1\right).\left(P\left(A/A_1\right)\right)}{P\left(A_1\right).P\left(A/A_1\right)+P\left(A_2\right).P\left(A/A_2\right)}$ (i)
In a pack of $52$ cards $26$ are red and $26$ are black.
Now $P\left(A_1\right)=$probability that the missing card is red$=\frac{{}^{26}{\text{C}}_1}{{}^{52}{\text{C}}_1}=\frac{26}{52}=\frac{1}{2}$
$P\left(A_2\right)=$probability that the missing card is black$=\frac{26}{52}=\frac{1}{2}$
$P\left(A/A_1\right)=$probability of drawing a red card when the missing card is red.
$=\frac{25}{51}$
[$\because$ Total number of cards left is $51$ out of which $25$ are red and $26$ are black as them is sing card is red]
Again $P\left(A/A_2\right)=$Probability of drawing a red card when the missing card is back$=\frac{26}{51}$
Now from (i), required probability, $P\left(A_1/A\right)=\frac{\frac{1}{2}.\frac{25}{51}}{\frac{1}{2}.\frac{25}{51}+\frac{1}{2}.\frac{26}{51}}=\frac{25}{51}$