Question

From a well-shuffled pack of $52$ cards four cards are drawn at random. Find the probability that all are of the same suit.

Answer 1

Hint: First we will write down what a deck of cards contain and then we will find the number of conditional cases by applying $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ for all the four suits then we will again apply this to find out the total number of cases. Finally, we will put it in the probability formula that is $\text{Probability}=\dfrac{\text{Number of conditional cases}}{\text{Total number of cases}}$ and get the required answer.

Complete step-by-step solution:
We have the total number of cards $=52$ and in each suit, the number of cards is $=13$ and there are a total of $4$ type of suits that are hearts, diamond, club, and spades.
Now we know that the probability of an event happening is as following:
$\text{Probability}=\dfrac{\text{Number of conditional cases}}{\text{Total number of cases}}$
Now let’s find out the total number of conditional cases, we are given the conditional cases is to select four cards and all of them should be of the same suit, it can be either heart or spades or clubs or diamonds.
So, the total number of conditional cases = No. of case of selecting all $4$ cards of hearts suit $+$ No. of the case of selecting all $4$ cards of spades suit $+$ No. of the case of selecting all $4$ cards of clubs suit $+$ No. of the case of selecting all $4$ cards of diamonds suit.
Now we know that when we have to select objects we use: $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ ,
Where $n$ is the total number of objects and $r$ is the number of objects which we have to select.
Now, there are $13$ cards in each suit and we have to choose $4$ out of them, therefore $n=13$ and $r=4$ , applying the above formula:
$^{13}{{C}_{4}}=\dfrac{13!}{4!\left( 13-4 \right)!}=\dfrac{13!}{4!9!}=715$
Therefore, for all the four suits we will have the following number of conditional cases = $^{13}{{C}_{4}}{{+}^{13}}{{C}_{4}}{{+}^{13}}{{C}_{4}}{{+}^{13}}{{C}_{4}}=715+715+715+715=2860$
Now, we will find the total number of cases, since there are total of $52$ cards and we have to select $4$ case, , therefore $n=52$ and $r=4$ , applying the selection formula : $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
\[^{52}{{C}_{4}}=\dfrac{52!}{4!\left( 52-4 \right)!}=~\dfrac{52!}{4!48!}\]
you can expand $52!$ as $52!=48!\times 49\times 50\times 51\times 52$ and cancel $48!$ Therefore,
\[\dfrac{52!}{4!48!}=\dfrac{48!\times 49\times 50\times 51\times 52}{4!48!}=\dfrac{49\times 50\times 51\times 52}{1\times 2\times 3\times 4}=270725\]

We will now apply, the formula for probability:
$\text{Probability}=\dfrac{\text{Number of conditional cases}}{\text{Total number of cases}}=\dfrac{2860}{270725}=0.0106$
Therefore the probability of getting all four cards of the same suit = $0.0106$

Note: Students can mistake in applying $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ as the number are bigger so the calculation might get messy so one needs to be careful there and also student may apply the standard formula for probability and can write it as $\dfrac{4}{52}$ or $\dfrac{13}{52}\times 4$, these mistakes must be avoided.