Question

Four cards are drawn at random from a pack of 52 cards. What is the probability of getting all four cards of the same suit?\[\]
A. $\dfrac{4}{65}$ \[\]
B. $\dfrac{4\left( ^{13}{{C}_{4}} \right)}{^{52}{{C}_{4}}}.$\[\]
C. $\dfrac{^{13}{{C}_{4}}}{^{52}{{C}_{4}}}.$\[\]
D. $\dfrac{4\left( ^{52}{{C}_{13}} \right)}{^{52}{{C}_{4}}}.$\[\]

Answer 1

Hint: Find out the total number of ways you can choose 4 cards from any particular suit (hearts, diamond, spade and club) . Find the number of ways you can choose any 4 cards from the pack. The ratio of number ways to select 4 cards from any particular suit to any card from the pack is the required probability. \[\]

Complete step-by-step answer:
We know that the selection of $r$ entities from $n$ unique entities is given by $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We know from definition of probability that if there is $n\left( A \right)$ number of ways of event $A$ occurring and $n\left( S \right)$ is the size of the sample space then the probability of the event $A$ occurring is $\dfrac{n\left( A \right)}{n\left( S \right)}$. \[\]

 The event is choosing 4 cards from the same suit. We know that in the pack of 52 cards which are not identical. There are 4 suits called hearts, diamond, spade and club. Each suit has $\dfrac{52}{4}=13$ cards. We can choose the 4 cards from the heart suit in using combinatorial formula $^{13}{{C}_{4}}$ ways. We can similarly choose 4 cards from the suits diamond, spade or club in $^{13}{{C}_{4}}$ ways. So we can choose 4 cards from any particular suit in $4\left( ^{13}{{C}_{4}} \right)$ ways. So $n\left( A \right)=4\left( ^{13}{{C}_{4}} \right)$\[\]
The sample space will be determined from the selection of any 4 cards. The number of ways we can select 4 cards from 52 cards is $^{52}{{C}_{4}}=n\left( S \right)$.
So the probability of choosing 4 cards the same suit is $\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{4\left( ^{13}{{C}_{4}} \right)}{^{52}{{C}_{4}}}.$\[\]
So the correct option is B. \[\]

Note: There are certain assumptions that are behind this problem. The cards cannot be replaced after selecting a card, the back of the cards identical and well-shuffled before selection and there is no extra card which is not from any suit like joker.