Question

Four cards are drawn at a time from a pack of 52 playing cards. Find the probability of getting all the four cards of the same suit.

• A. $\frac{4}{65}$
• B. $\frac{4\times {13}_{C_4}}{{52}_{C_4}}$
• C. $\frac{{13}_{C_4}}{{52}_{C_4}}$
• D. $\frac{4\times {52}_{C_{13}}}{{52}_{C_4}}$

Answer 1

Answer: (B)

$\frac{4\times {13}_{C_4}}{{52}_{C_4}}$

Since $4$ cards can be drawn at a time from a pack of $52$ cards in ${}^{52}{C}_4$ ways,
total number of elementary events ${=}^{52}{C}_4$
Consider the following events:
$A=$ Getting all spade cards;
$B=$ Getting all club cards;
$C=$ Getting all diamonds cards and
$D=$ Getting all heart cards.
Then $A,B,C,D$ are mutually exclusive events such that
$P\left(A\right)=\frac{{}^{13}{C}_4}{{}^{52}{C}_4},P\left(B\right)=\frac{{}^{13}{C}_4}{{}^{52}{C}_4},P\left(C\right)=\frac{{}^{13}{C}_4}{{}^{52}{C}_4}$ and $P\left(D\right)=\frac{{}^{13}{C}_4}{{}^{52}{C}_4}$

Now, required probability $=P(A\cup B\cup C\cup D)$
$=P\left(A\right)+P\left(B\right)+P\left(C\right)+P\left(D\right)$ (bu addition theorem)
$=4.\frac{{}^{13}{C}_4}{{}^{52}{C}_4}$