Question

you are dealt a hand of 6 cards from a standard deck of 52 cards. find the probability of being dealt three clubs and three cards with one card of each other remaining suit. the probability is . (round to six decimal places as needed.)

Explanation:

Step1: Calculate total number of ways to choose 6 - card hands

The number of ways to choose 6 cards from a deck of 52 cards is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 52$ and $r=6$. So, $C(52,6)=\frac{52!}{6!(52 - 6)!}=\frac{52!}{6!46!}=\frac{52\times51\times50\times49\times48\times47}{6\times5\times4\times3\times2\times1}=20358520$.

Step2: Calculate number of favorable hands

We want 3 clubs and 3 non - clubs. There are 13 clubs in a deck and $52-13 = 39$ non - clubs.
The number of ways to choose 3 clubs from 13 is $C(13,3)=\frac{13!}{3!(13 - 3)!}=\frac{13!}{3!10!}=\frac{13\times12\times11}{3\times2\times1}=286$.
The number of ways to choose 3 non - clubs from 39 is $C(39,3)=\frac{39!}{3!(39 - 3)!}=\frac{39!}{3!36!}=\frac{39\times38\times37}{3\times2\times1}=9139$.
The number of favorable hands (3 clubs and 3 non - clubs) is the product of the number of ways to choose 3 clubs and the number of ways to choose 3 non - clubs, so $C(13,3)\times C(39,3)=286\times9139 = 2613754$.

Step3: Calculate the probability

The probability $P$ is the number of favorable hands divided by the total number of hands. So, $P=\frac{C(13,3)\times C(39,3)}{C(52,6)}=\frac{2613754}{20358520}\approx0.128385$.

Answer:

$0.128385$