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February 27, 2015

February 27, 2015

Posted by **SNOOPY** on Thursday, April 17, 2008 at 11:24pm.

- PROBABILITY -
**PsyDAG**, Saturday, April 19, 2008 at 11:27amThe probability ofdrawing

*just*one club in the 52 draws = 1/4 * (3/4)^51

The probability of drawing two clubs — which fits the "at least" description — in 52 draws = (1/4)^2 * (3/4)^50

The probability of drawing three clubs in 52 draws = (1/4)^3 * (3/4)^49

Or you could go from the opposite direction, 1 - probability of drawing no clubs.

Does that lead you to a formula?

I hope this helps. Thanks for asking.

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