Question

A and B draw two cards each, one after another, from a well-shuffled pack of 52 cards. The probability that all the four cards drawn are of the same suit is.
A. $\dfrac{44}{85\times 49}$
B. $\dfrac{11}{85\times 49}$
C. $\dfrac{13\times 24}{17\times 25\times 49}$
D. None of these

Answer 1

Hint: We will find the probability of picking 4 cards from 13 cards 4 times because in a full deck of 52 cards there are 4 suits, they are: spade, diamond, clubs and hearts. Then we will divide the favorable outcome by total outcomes.

Complete step-by-step answer:
It is given in the question that A and B draw 2 cards each from a well-shuffled pack of 52 cards. Then, we have to find the probability that all the 4 cards drawn by A and B are from the same suit.
We know that if we have ‘n’ number of things and we have to pick ‘r’ from ‘n’ then the number of ways of picking ‘r’ from ‘n’ is given by ${}^{n}{{C}_{r}}$. And also ${}^{n}{{C}_{r}}$ is expanded as, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. So, we have to select 4 cards from a well-shuffled deck of 52 cards. Then, the number of ways of picking 4 cards from well-shuffled 52 cards is given by – ${}^{52}{{C}_{4}}$. Now, we have to find the probability that all the 4 cards are drawn from the same suit.

We know that in a well-shuffled 52 cards there are 4 suits each with 13 cards. So, the number of ways of picking 4 cards from 13 spades is ${}^{13}{{C}_{4}}$.
Similarly, the number of ways of picking 4 cards from diamond is ${}^{13}{{C}_{4}}$. Now, the number of ways of picking 4 cards from clubs and hearts is also ${}^{13}{{C}_{4}}$. So, we have the total number of ways as –
$=$ Number of ways of picking (Spades $\times $ Diamond $\times $ Club $\times $ Heart)
$={}^{13}{{C}_{4}}\times {}^{13}{{C}_{4}}\times {}^{13}{{C}_{4}}\times {}^{13}{{C}_{4}}$
$=4\times {}^{13}{{C}_{4}}$
We know that probability = number of favoured outcome ÷ number of total outcome. Therefore, we can substitute the value as,
$=\dfrac{4\times {}^{13}{{C}_{4}}}{{}^{52}{{C}_{4}}}$
$=\dfrac{4\times \dfrac{13!}{4!\times 9!}}{\dfrac{52!}{4!\times 48!}}$
$=\dfrac{\dfrac{4\times 13\times 12\times 11\times 10\times 9!}{4!\times 9!}}{\dfrac{52\times 51\times 50\times 49\times 48!}{4!\times 48!}}$
Cancelling the similar terms from numerator and denominator, we get
$=\dfrac{4\times 13\times 12\times 11\times 10}{52\times 51\times 50\times 49}$
$=\dfrac{44\times 1560}{132600\times 49}$
$=\dfrac{44}{85\times 49}$
So, the probability of picking 4 cards from a well-shuffled deck of 52 cards from the same suit is $\dfrac{44}{85\times 49}$. Option (A) is the correct answer.

Note: Use the formula of ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ wherever required in the solution. But, remember that ${}^{n}{{C}_{r}}$ and ${}^{n}{{P}_{r}}$ look alike but they are different. The expansion of ${}^{n}{{C}_{r}}$ is given by $\dfrac{n!}{r!\left( n-r \right)!}$ and the expansion of ${}^{n}{{P}_{r}}$ is given by $\dfrac{n!}{\left( n-r \right)!}$.