A deck with 2 missing cards . A card is drawn .

combinatoricsconditional probabilitypermutationsprobabilityprobability distributions

From a deck of cards , 2 cards are missing . Now , a
Card is Drawn from this deck . Find the probability of this card being a king .

Now how I’ve gone about is ,
Since 2 cards are drawn , there can be 3 cases involving a king in those 2 cards .
Either 2 kings , 1 king or no king as a part of those 2 missing cards .
Using these Ive formed cases using conditional probability, but I’m not able to go on ahead .

Best Answer

The probability you are asking for is:

probability that $0$ kings are missing $*$ probability to chose a king given that $0$ kings are missing

plus

probability that $1$ king is missing $*$ probability to chose a king given that $1$ king is missing

plus

probability that $2$ kings are missing $*$ probability to chose a king given that $2$ kings are missing

So we have:

$P(king)=\frac{48*47}{52*51}*\frac{4}{50}+\frac{\begin{pmatrix} 48 \\ 1 \\ \end{pmatrix}*\begin{pmatrix} 4 \\ 1 \\ \end{pmatrix}}{\begin{pmatrix} 52 \\ 2 \\ \end{pmatrix}}*\frac{3}{50}+\frac{4*3}{52*51}*\frac{2}{50}=\frac{48*47}{52*51}*\frac{4}{50}+\frac{48*4}{1326}*\frac{3}{50}+\frac{4*3}{52*51}*\frac{2}{50}\simeq 0.076923$

which is actually equal to the probability of choosing a king from a normal deck(without 2 removed cards). That is because you can view the same question as:

Chose three cards from a normal deck. What is the probability of the third being a king? Of course $4/52\simeq 0.076923$.

Note: This works if the missing cards were taken each with the same probability