Question

you are dealt a hand of 8 cards from a standard deck of 52 cards. find the probability of being dealt five 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 the total number of ways to choose 8 cards from 52

The formula for combinations is \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 52\) and \(r=8\).

$$C(52,8)=\frac{52!}{8!(52 - 8)!}=\frac{52!}{8!44!}=\frac{52\times51\times50\times49\times48\times47\times46\times45}{8\times7\times6\times5\times4\times3\times2\times1}=752538150$$

Step2: Calculate the number of ways to get 5 clubs and 3 non - club cards (one from each of the other 3 suits)

• There are 13 clubs in a deck. The number of ways to choose 5 clubs is \(C(13,5)=\frac{13!}{5!(13 - 5)!}=\frac{13\times12\times11\times10\times9}{5\times4\times3\times2\times1}=1287\)
• For the non - club cards (3 suits, 1 card from each suit). There are 13 cards in each non - club suit. The number of ways to choose 1 card from each of the 3 non - club suits is \(C(13,1)\times C(13,1)\times C(13,1)=13\times13\times13 = 2197\)

• The number of favorable hands is \(N = C(13,5)\times13\times13\times13=1287\times2197 = 2827539\)

Step3: Calculate the probability

The probability \(P=\frac{\text{Number of favorable hands}}{\text{Total number of hands}}\)

$$P=\frac{2827539}{752538150}\approx0.003757$$

Answer:

\(0.003757\)