Question

A $5$-card hand is dealt from a well-shuffled deck of $52$playing cards. What is the probability that the hand contains at least one card from each of the four suits?

Short answer

$P\left(A\right)=\frac{4·\left(\begin{array}{c}13\\ 2\end{array}\right)·{13}^3}{\left(\begin{array}{c}52\\ 5\end{array}\right)}\approx 0.264$

Choose the suits that appear twice, and then all the denominations.

Step-by-step solution

Given Information.

A $5$-card hand is dealt from a well-shuffled deck of $52$playing cards.

Explanation.

Experiment: Draw $5$random cards from a $52$card deck.

What is the probability that all $4$suits appear in the $5$drawn cards?

Outcome space $S$contains every combination of $5$cards$52$.

All events $S$are considered equally likely, the probability of an event $A\subseteq S$is:

$P\left(A\right)=\frac{\left|A\right|}{\left|S\right|}$

where$\left|X\right|$denotes the number of elements in$X$.

It is shown that the number of $5$card combinations of $52$different cards is$\left(\begin{array}{c}52\\ 5\end{array}\right)=\left|S\right|$.

Event$A$- all $4$suits appear in the $5$cards.

Count the number of sample events$A$using the basic principle of counting.

If all four suits appear in the five cards, there will be two cards from some suit in the five cards, and one card from each of the other three.

Choose the suit that appears twice - $4$ways.

Choose two cards from $13$in that suit -$\left(\begin{array}{c}13\\ 2\end{array}\right)$ways.

Choose one of the $13$cards in each of the remaining three suits $-{13}^3$ways.

Those three choices do not affect each other in the number of ways they can end, so the number of events$A$is$\left|A\right|=4·\left(\begin{array}{c}13\\ 2\end{array}\right)·{13}^3$.

$P\left(A\right)=\frac{4·\left(\begin{array}{c}13\\ 2\end{array}\right)·{13}^3}{\left(\begin{array}{c}52\\ 5\end{array}\right)}\approx 0.264$