Question

What is the probability of getting a “FULL HOUSE” in five cards drawn in a poker game from a standard pack of $52$-cards?

[A FULL HOUSE consists of $3$ cards of the same kind (eg, $3$ Kings) and $2$ cards of another kind (eg, $2$ Aces)]

• A. $\frac{6}{4165}$
• B. $\frac{4}{4165}$
• C. $\frac{3}{4165}$
• D. None of these

Answer 1

The correct option is A

$\frac{6}{4165}$

Explanation of correct option:

Step 1: Calculate the total number of outcomes:

The formula of combination is ${}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{=}\frac{n!}{r!\left(n-r\right)!}$

The number of ways of selecting $5$ cards from $52$ cards is:

${}^{52}C_5=\frac{52!}{5!47!}$

$\Rightarrow$ ${}^{52}C_5=2598960$

Step 2: Calculate the number of favorable outcomes:

The number of ways of selecting $2$ cards of the same kind from the $4$ cards in the deck is,

${}^{4}C_2=6$

There are $13$ different kinds of cards.

So the total number of combinations possible of $2$ cards is: $6\times 13=78$

The number of ways of selecting $3$ cards of the same kind from 4 cards is:

${}^{4}C_3=4$

Since the $3$-of-a-kind suit must be different from the $2$-of-a-kind suit, the possible combination of this is: $4\times 12=48$

So, the total number of ways is: $48\times 78=3744$

Step 3: Finding the required probability

Hence, the probability of getting a “FULL HOUSE” in five cards drawn in a poker game from a standard pack of $52$-cards is,

$P=\frac{3744}{2598960}$

$\Rightarrow$ $P=\frac{6}{4165}$

Hence, the correct option is $\left(A\right)$.