What is the probability of getting a "FULL HOUSE" in five cards drawn in a poker game from a standard pack of $$52$$ -cards ?
[A "FULL HOUSE" consists of $$3$$ cards of the same kind (eg, $$3$$ Kings) and $$2$$ cards of another kind (eg, $$2$$ Aces)]
A.
$$\frac{6}{{4165}}$$
B.
$$\frac{4}{{4165}}$$
C.
$$\frac{3}{{4165}}$$
D.
none of these
Answer :
$$\frac{6}{{4165}}$$
Solution :
There are $$6$$ ways to select $$2$$ cards of the same kind from the $$4$$ cards in the deck and there are $$13$$ different kinds of cards, so the total number of combinations possible of $$2$$ cards is $$6 \times 13 = 78.$$
There are $$4$$ ways to choose $$3$$ cards of the same kind from $$4$$ cards of the same kind, but because the $$3$$-of-a-kind suit must be different from the $$2$$-of-a-kind suit, the possible combinations of this is, $$4 \times 12 = 48.$$
So total number of ways is $$ = 48 \times 78 = 3744.$$
Sample space is $${}^{52}{C_5} = 2598560.$$
Required Probability $$ = \frac{{3744}}{{2598560}} = \frac{6}{{4165}}.$$
Releted MCQ Question on Statistics and Probability >> Probability
Releted Question 1
Two fair dice are tossed. Let $$x$$ be the event that the first die shows an even number and $$y$$ be the event that the second die shows an odd number. The two events $$x$$ and $$y$$ are:
Two events $$A$$ and $$B$$ have probabilities 0.25 and 0.50 respectively. The probability that both $$A$$ and $$B$$ occur simultaneously is 0.14. Then the probability that neither $$A$$ nor $$B$$ occurs is
The probability that an event $$A$$ happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event $$A$$ happens at least once is
If $$A$$ and $$B$$ are two events such that $$P(A) > 0,$$ and $$P\left( B \right) \ne 1,$$ then $$P\left( {\frac{{\overline A }}{{\overline B }}} \right)$$ is equal to
(Here $$\overline A$$ and $$\overline B$$ are complements of $$A$$ and $$B$$ respectively).
A.
$$1 - P\left( {\frac{A}{B}} \right)$$
B.
$$1 - P\left( {\frac{{\overline A }}{B}} \right)$$
C.
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
D.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$