Please Help: Counting/Probability

[Aria1]

Conditional Probability

Drawing a 5-card hand from a standard deck of 52 cards: Let X be a random variable representing the number of different suits in your hand. Let K be the random variable representing how many kings are in your hand. Given that there is one king, find the conditional probability mass function of X.

For example, Pr{X = 2 | K=1} = Pr{ X=2 and K=1}/Pr{K=1} This must be done for X=1, X=2, X=3, and X=4. I have calculated Pr{K=1} as (4C1)(48C4)/(52C5), but am having trouble with the joint probability of kings and suits.

Please help, due tomorrow!!!

 

[thepillow]

Drawing a 5-card hand from a standard deck of 52 cards: Let X be a random variable representing the number of different suits in your hand. Let K be the random variable representing how many kings are in your hand. Given that there is one king, find the conditional probability mass function of X.

For example, Pr{X = 2 | K=1} = Pr{ X=2 and K=1}/Pr{K=1} This must be done for X=1, X=2, X=3, and X=4. I have calculated Pr{K=1} as (4C1)(48C4)/(52C5), but am having trouble with the joint probability of kings and suits.

Please help, due tomorrow!!!

Sorry for the late response. This is a pretty interesting problem though. Definitely not easy!

Here's a way to find the number of 2 suit hands that contain exactly 1 king, which is {X=2 and K=1}, and a similar approach can be applied to 1 suit, 3 suit and 4 suit hands). A 5 card hand with exactly 2 suits must have 3 cards from 1 suit and 2 cards from another suit. The 1 king in the hand could come from either of the suits, so we have:

Case 1: the king is from the suit with 3 cards
Case 2: the king is from the suit of 2 cards


CASE 1​

\begin{align*}
\text{Pick 2 suits from 4}&= \binom{4}{2} \\
\text{Pick which is suit with 3 of the 5 cards} &= \binom{2}{1} \\
\text{Pick king and any 2 others from chosen suit} &= \binom{1}{1}\binom{12}{2} \\
\text{Pick any 2 cards from other suit} &= \binom{13}{2}
\end{align*}

CASE 2​

\begin{align*}
\text{Pick 2 suits from 4}&= \binom{4}{2} \\
\text{Pick which is suit with 3 of the 5 cards} &= \binom{2}{1} \\
\text{Pick and 3 cards from chosen suit} &=\binom{13}{3} \\
\text{Pick king and any 1 other from other suit} &= \binom{1}{1}\binom{12}{1}
\end{align*}

So the probability this happens, i.e. P(X = 2 and K = 1) = (Case 1 + Case 2) divided by the total number of 5 card hands (52 choose 5).


Does that give you enough to go on to find the necessary probabilities?