Solveeit Logo
Login

Question

Question: Four cards are drawn at random from a pack of 52 cards. What is the probability of getting all four ...

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}}}.

Explanation

Solution

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 rr entities from nn unique entities is given by nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}. We know from definition of probability that if there is n(A)n\left( A \right) number of ways of event AA occurring and n(S)n\left( S \right) is the size of the sample space then the probability of the event AA occurring is n(A)n(S)\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 524=13\dfrac{52}{4}=13 cards. We can choose the 4 cards from the heart suit in using combinatorial formula 13C4^{13}{{C}_{4}} ways. We can similarly choose 4 cards from the suits diamond, spade or club in 13C4^{13}{{C}_{4}} ways. So we can choose 4 cards from any particular suit in 4(13C4)4\left( ^{13}{{C}_{4}} \right) ways. So n\left( A \right)=4\left( ^{13}{{C}_{4}} \right)$$$$$ The sample space will be determined from the selection of any 4 cards. The number of ways we can select 4 cards from 52 cards is ^{52}{{C}{4}}=n\left( S \right).Sotheprobabilityofchoosing4cardsthesamesuitis. So the probability of choosing 4 cards the same suit is \dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{4\left( ^{13}{{C}{4}} \right)}{^{52}{{C}_{4}}}.$$$$$
So the correct option is B. $$$$

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.