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: [tex]^{52}C_5=\frac{52!}{5!47!}[/tex]
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:
[tex]^{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!}[/tex]
Thus, the required probability is:
[tex]\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!}}[/tex]
[tex]=\frac{3744}{2598960}\\\\\approx0.00144[/tex]
Hence, the required probability is 0.00144 or 0.144%.
