Question

Compute the probability of being dealt a full house on a initial deal of $5$ cards.

• A. $\frac{{}^{13}{C}_1{}^4{C}_3{}^{12}{C}_1}{{}^{52}{C}_5}$
• B. $\frac{{}^{13}{C}_1{}^{12}{C}_1{}^4{C}_2}{{}^{52}{C}_5}$
• C. $\frac{{}^{13}{C}_1{}^4{C}_3{}^{12}{C}_1{}^4{C}_2}{{}^{52}{C}_5}$
• D. $\frac{{}^{13}{C}_1{}^4{C}_3{}^4{C}_2}{{}^{52}{C}_5}$

Answer 1

The correct option is C $\frac{{}^{13}{C}_1{}^4{C}_3{}^{12}{C}_1{}^4{C}_2}{{}^{52}{C}_5}$

Full house $3$ cards has same numbers and $2$ cards has same numbers.

Example: $3k^{\prime }s$ and $2n^{\prime }s$

$3$ cards should have same numbers from $13cards$$={}^{13}{C}_1$

$A$ cards will be of same number

We have to select $3$ from $4$ cards $={}^4{C}_3$

$\Rightarrow {}^{13}{C}_1\times {}^4{C}_3$

Remaining $2$cards should be same number from remaining $12$ numbers $={}^{12}{C}_1$

These numbers will be on $4$ cards

We have selected two cards$={}^4{C}_2$

$={}^{12}{C}_1\times {}^4{C}_2$

No of ways of selecting $5$ cards $={}^{52}{C}_5$

Total probability $=\frac{\text{no of ways of selecting full house}}{\text{no of ways of selecting 5 cards}}$

$=\frac{{}^{13}{C}_1\times {}^4{C_3\times }^{12}{C}_1\times {}^4{C}_2}{{}^{52}{C}_5}$