Question

A card from a pack of 52 cards is lost. From the remaining cards of the pack two cards are drawn at random and are found to be spades.The probability of the missing card being a spade is P.Then 50p is equal to

Answer 1

Let $E_1$ and $E_2$ be the respective events of choosing a diamond card and a card which is not diamond.
Let A denote the lost card.
out of $52$ cards, $13$ cards are diamond and $39$ cards are not diamond.
$\therefore P\left(E_1\right)=\frac{13}{52}=\frac{1}{4}$
and $P\left(E_2\right)=\frac{39}{52}=\frac{3}{4}$
when one diamond card is lost , there are $12$ diamond cards out of $51$ cards.
Two cards can be drawn out of $12$ cards in ${}^{12}{C}_2$ ways.
Similarly, two diamond cards can be drawn out of $51$ cards in ${}^{51}{C}_2$ ways.
the probability of getting two cards, when one diamond card is lost is given by $P\left(\frac{A}{E_1}\right)$
$P\left(\frac{A}{E_1}\right)=\frac{{}^{12}{C}_2}{{}^{51}{C}_2}=\frac{12!}{2!\times 10!}\times \frac{2!\times 49!}{51!}=\frac{11\times 12}{50\times 5}=\frac{22}{425}$.
When the lost card is not a diamond, there are $13$ diamond cards out of $51$ cards. two cards can be drawn out of $13$ diamond cards in ${}^{13}{C}_2$ways, whereas two cards can be drawn out of $51$ cards in ${}^{51}{C}_2$ ways.
The probability of getting two cards, when one card is lost which is not diamond, is given by $P\left(\frac{A}{E_2}\right)$.
$P\left(\frac{A}{E_2}\right)=P\left(\frac{A}{E_2}\right)=\frac{{}^{13}{C}_2}{{}^{51}{C}_2}=\frac{13!}{2!\times 11!}\times \frac{2!\times 49!}{51!}=\frac{12\times 13}{50\times 51}=\frac{26}{425}$
the probability that the lost card is diamond by $P\left(\frac{E_1}{A}\right)$
by using Baye's Theorem,
$P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right).P\left(\frac{A}{E_1}\right)}{P\left(E_1\right).P\left(\frac{A}{E_1}\right)+P\left(E_2\right).P\left(\frac{A}{E_2}\right)}=\frac{\frac{1}{4}.\frac{22}{425}}{\frac{1}{4}.\frac{22}{425}+\frac{3}{4}.\frac{26}{425}}$
$==\frac{\frac{1}{425}\left(\frac{22}{4}\right)}{\frac{1}{425}(\frac{22}{4}+\frac{26\times 3}{4})}=\frac{\frac{11}{2}}{25}=\frac{11}{50}$