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,

Hint: Find the probability of occurrence of two spade cards and the two cards which are not spades separately. You have to do it twice, first considering the case that lost card is a spade card and second considering the case that lost card is not a spade card.

After that, use Baye’s theorem to find the probability of the missing card being a spade card.

Complete step-by-step answer:

Let \[{{E}_{1}}=\]Event of choosing a spade card

\[{{E}_{2}}=\]Event of choosing a card which is not spade.

Let, A denote the lost card.

Out of 52 cards of the deck, there are 13 spade cards.

Hence, there are $\left( 52-13 \right)=39$cards, which are not spades.

\[\begin{align}

  & \therefore P\left( {{E}_{1}} \right)=\dfrac{Number\ of\ favourable\ outcomes}{Total\

number\ ofpossible\ outcomes} \\

 & P\left( {{E}_{1}} \right)=\dfrac{Number\ of\ Spade\ cards\ in\ the\ pack}{Total\ number\

ofcards\ in\ the\ pack} \\

 & P\left( {{E}_{1}} \right)=\dfrac{13}{52} \\

 & P\left( {{E}_{1}} \right)=\dfrac{1}{4} \\

 & \therefore P\left( {{E}_{2}} \right)=\dfrac{Number\ of\ favourable\ outcomes}{Total\

number\ ofoutcomes} \\

 & P\left( {{E}_{2}} \right)=\dfrac{39}{52} \\

 & P\left( {{E}_{2}} \right)=\dfrac{3}{4} \\

\end{align}\]

If one spade card is lost, there are 12 spade cards out of the remaining 51 cards.

According to the question, two cards are drawn from the remaining cards and found to be spades.

Now, two cards can be drawn out of 12 spades cards in ${}^{12}{{C}_{2}}$ ways.

(Since, selection of ‘r’ things, from ‘n’ things can be done in ${}^{n}{{C}_{r}}$ways.)

Similarly, two spade cards can be drawn out of 51 cards in ${}^{51}{{C}_{2}}$ ways.

The probability of getting two spade cards, when one spade card is lost is given by,

$\begin{align}

  & P\left( \dfrac{A}{{{E}_{1}}} \right)=\dfrac{{}^{12}{{C}_{2}}}{{}^{51}{{C}_{2}}} \\

 & \because {}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!} \\

 & P\left( \dfrac{A}{{{E}_{1}}} \right)=\dfrac{\dfrac{12!}{10!2!}}{\dfrac{51!}{49!2!}} \\

 & \Rightarrow \dfrac{12!}{10!2!}\times \dfrac{2!49!}{51!} \\

 & \Rightarrow \dfrac{12\times 11}{2}\times \dfrac{2\times 1}{51\times 50} \\

 & \Rightarrow \dfrac{12\times 11}{50\times 51}=\dfrac{22}{425} \\

\end{align}$

Now, consider the case when the lost card is not a spade, there are 13 spade cards out of 51 remaining cards. Two cards can be drawn out of 13 spade 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 spade is given by $P\left( \dfrac{A}{{{E}_{2}}} \right)$$\begin{align}

  & P\left( \dfrac{A}{{{E}_{2}}} \right)=\dfrac{{}^{13}{{C}_{2}}}{{}^{51}{{C}_{2}}} \\

 & \Rightarrow \dfrac{\dfrac{13!}{11!2!}}{\dfrac{51!}{49!2!}} \\

 & \Rightarrow \dfrac{13\times 12}{2}\times \dfrac{2\times 1}{51\times 50} \\

 & \Rightarrow \dfrac{13\times 12}{51\times 50} \\

 & \Rightarrow \dfrac{26}{425} \\

\end{align}$

The probability that the lost card is diamond by $P\left( \dfrac{{{E}_{1}}}{A} \right)$.

By using Baye’s theorem,

$\begin{align}

  & P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{P\left( {{E}_{1}} \right).P\left( \dfrac{A}{{{E}_{1}}}

\right)}{P\left( {{E}_{1}} \right).P\left( \dfrac{A}{{{E}_{1}}} \right)+P\left( {{E}_{2}}

\right).P\left( \dfrac{A}{{{E}_{2}}} \right)}...............\left( 1 \right) \\

 & Here, \\

 & P\left( {{E}_{1}} \right)=\dfrac{1}{4} \\

 & P\left( \dfrac{A}{{{E}_{1}}} \right)=\dfrac{22}{425} \\

 & P\left( {{E}_{2}} \right)=\dfrac{3}{4} \\

 & P\left( \dfrac{A}{{{E}_{2}}} \right)=\dfrac{26}{425} \\

\end{align}$

Putting these values in equation (1), we get,

\[\begin{align}

  & P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{\dfrac{1}{4}\times \dfrac{22}{425}}{\dfrac{1}{4}\times \dfrac{22}{425}+\dfrac{3}{4}\times \dfrac{26}{425}} \\

 & \Rightarrow \dfrac{\dfrac{1}{425}\left( \dfrac{22}{4} \right)}{\dfrac{1}{425}\left( \dfrac{22}{4}+\dfrac{26\times 3}{4} \right)}=\dfrac{\dfrac{11}{2}}{25}=\dfrac{11}{50} \\

\end{align}\]

Now, $P\left( \dfrac{{{E}_{1}}}{A} \right)=required\ probability=\dfrac{11}{50}=P$

$\therefore 50P=\dfrac{11}{50}\times 50=11$


Note: Students can make mistakes by not considering two cases, where the first case is to consider the lost card to be spade, then calculate the probability of selection of two spade cards. Now, consider the second case, that the lost card is not spade and then calculate the probability of selection of two spade cards.