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2 cards are drawn from a 52 card deck. The random variable X represents the number of aces drawn. The random variable distribution:

X = 0 : P(0) = 0.849

X = 1 : P(1) = 0.145

X = 2 : P(2) = 0.005

Is this how you do it?

Probability of the draws:

0 aces out of 2 cards:

C(48,2) * C(4,0) / C(52,2)

1128 / 1328 = 0.849

1 ace out of 2 cards:

C(48,1) * C(4,1) / C(52,2)

48 * 4 / 1328 = 0.145

2 aces out of 2 cards:

C(48,0) * C(4,2) / C(52,2)

6 / 1328 = 0.005

user395040
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  • The probabilities should add up to 1, and yours add up only to 1326/1328. – pixel Dec 03 '16 at 23:54
  • So just a rounding error? But my probability calculations are sound? – user395040 Dec 04 '16 at 00:03
  • It doesn't seem like a rounding error, since you are showing fractions (not decimals) that don't add up to 1. So you may want to re-check your numbers along the lines of the answer given below. – pixel Dec 04 '16 at 13:05

1 Answers1

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The probability of $X=x$ is given by

$$\Pr(X=x)=\frac{\binom{4}{x}\binom{52-4}{2-x}}{\binom{52}{2}}$$

where $X\in[0,2]$.

This gives: $$\Pr(X=0)=\frac{188}{221}\approx 0.8507$$ $$\Pr(X=1)=\frac{32}{221}\approx 0.1448$$ $$\Pr(X=2)=\frac{1}{221}\approx 0.0045$$

msm
  • 7,147