Q. 2.5 - A First Course in Probability

Question

An ordinary deck of$52$ cards is shuffled. What is the probability that the top four cards have

(a) different denominations?

(b) different suits?

Step-by-Step Solution

Answer

$a)$$P($the four cards have different denominations$)=\frac{\left(\begin{array}{c}13\\ 4\end{array}\right)·4^4}{\left(\begin{array}{c}52\\ 4\end{array}\right)}\approx 0.67611$.

$b) P($the four cards are of different suits$)=\frac{{13}^4}{\left(\begin{array}{c}52\\ 4\end{array}\right)}\approx 0.1055$.

Step 1 Given Information.

An ordinary deck of$52$ cards is shuffled.

Step 2 Explanation.

The described experiment is equivalent to:

Experiment: Four random cards are drawn from a standard $52$card deck.

Outcome space$S$ contains every combination of cards.

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

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

In the chapter$1.4$. it is shown that the number of four card combinations of$52$ different cards is$\left(\begin{array}{c}52\\ 4\end{array}\right)=\left|S\right|$.

Step 3 Part (a) Explanation.

$P($the four cards have different denominations)

Count all possible events from $S$where four cards have different denominations, using the basic principle (the product principle)

Then each of the cards can be in any of the four colors. Total of $4^4$different possibilities.

There are $\left(\begin{array}{c}13\\ 4\end{array}\right)·4^4$elements in the event where all denominations of cards are different.

$P\left(\text{ the four cards have different denominations }\right)=\frac{\left(\begin{array}{c}13\\ 4\end{array}\right)·4^4}{\left(\begin{array}{c}52\\ 4\end{array}\right)}\approx 0.67611$

Step 4 Part (b) Explanation.

$P$(the four cards are of different suits)

Count all possible events from $S$where four cards are of different suits, using the basic principle (the product principle)

First choose the four suits, in one way.

Then each of the cards in some suit can be in any of the thirteen cards. Total of ${13}^4$different possibilities.

There are ${13}^4$elements in the event where all denominations of cards are different.

$P\left(\text{ the four cards are of different suits }\right)=\frac{{13}^4}{\left(\begin{array}{c}52\\ 4\end{array}\right)}\approx 0.1055$