What is the probability of getting a hand of two clubs and three diamonds?

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What is the probability of getting a hand of two clubs and three diamonds?

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Answer 1 · 2015-11-20T14:15:45

$(\frac{13}{52} \times \frac{12}{51} \times \frac{13}{50} \times \frac{12}{49} \times \frac{11}{48}) \times^{5} C_{2} = \frac{267696}{31187520}$ $(13/52xx12/51xx13/50xx12/49xx11/48)xx""^5C_2 = 267696/31187520$

$\approx .008583433373349339$ $~~ .008583433373349339$

That is about $1$ $1$ in $116$ $116$

Explanation:

The probability of being dealt two clubs then three diamonds is:

$\frac{13}{52} \times \frac{12}{51} \times \frac{13}{50} \times \frac{12}{49} \times \frac{11}{48}$ $13/52xx12/51xx13/50xx12/49xx11/48$

But we don't mind what order we get these cards, so this probability needs to be multiplied by $^{5} C_{2} = \frac{5!}{2! 3!} = \frac{5 \times 4}{2} = 10$ $""^5C_2 = (5!)/(2!3!) = (5xx4)/2 = 10$ to represent the number of possible orders of clubs and diamonds.

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What is the probability of getting a hand of two clubs and three diamonds?

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