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- Jan 26, 2012

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Thread starter
- Admin
- #1

- Jan 26, 2012

- 4,055

--------------------

Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter
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- #2

- Jan 26, 2012

- 4,055

Solution:

There are 4 Queens from which to choose 3 and 43 cards left over from which to choose the other 2 cards. However these last two cards also cannot be the same, so for the first card we have 43 choices and for the second card we have 39 choices.

\(\displaystyle P[\text{3 Queens}]=\frac{\binom{4}{3} \left(\frac{43 \times 39}{2!}\right)}{\binom{47}{5}} \approx 0.0021865\)

\(\displaystyle P[\text{3 Queens}] \times 3 = \boxed{0.00656}\)

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