Question

Five cards are drawn from an ordinary deck of 52 playing cards. What is the probability that the hand drawn is a full house? (A full house is a hand that consists of two of one kind and three of another kind.)

Answer 1

Answer:

The required probability is 0.00144 or 0.144%.

Step-by-step explanation:

Consider the provided information.

Five cards are drawn from an ordinary deck of 52 playing cards.

A full house is a hand that consists of two of one kind and three of another kind.

The total number of ways to draw 5 cards are: $^{52}C_5=\frac{52!}{5!47!}$

Now we want two of one kind and three of another.

Let the hand has the pattern AAABB, where A and B are from distinct kinds. The number of such hands are:

$^{13}C_1\times^{4}C_3\times^{12}C_1\times^{4}C_2=\frac{13!}{12!}\times\frac{4!}{3!}\times\frac{12!}{11!}\times\frac{4!}{2!2!}$

Thus, the required probability is:

$\frac{^{13}C_1\times^{4}C_3\times^{12}C_1\times^{4}C_2}{^{52}C_5}=\frac{\frac{13!}{12!}\times\frac{4!}{3!}\times\frac{12!}{11!}\times\frac{4!}{2!2!}}{\frac{52!}{5!47!}}$

$=\frac{3744}{2598960}\\\\\approx0.00144$

Hence, the required probability is 0.00144 or 0.144%.