1. ## Random variable distribution

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

100%

3. ## Re: Random variable distribution

Your "formula" is correct, your arithmetic is not: $_{50}C_{2}= \frac{52!}{50!2!}= \frac{52(51)}{2}= 26(52)= 1326$, not 1328.

How I would do this: there are 52 cards in the deck of which 4 are aces and 48 are not. The probability the first card drawn is NOT an ace is 48/52= 12/13. Then there are 51 cards left of which 47 are not aces. The probability the second card drawn is also not an ace is 47/51. The probability neither card is an ace is $P(0)= \frac{12}{13}\frac{47}{51}= 0.8507$. P(0)= 0.8507.

The probability the first card is an ace is 4/52= 1/13. Then there are 51 cards left of which the 48 are not aces. The probability the second card is not an ace is 48/51. The probability the first card drawn is an ace and the second is not is $\frac{1}{13}\frac{48}{51}= 0.072$.
The probability the first card is not an ace is 48/52= 12/13. Then there are 51 cards left of which the 4 are aces. The probability the second card is an ace is 4/51. The probability the first card drawn is an ace and the second is not is $\frac{12}{13}\frac{4}{51}= 0.072$. (It is always true that these are the same- we are just changing the order of numbers in the numerator and denominator.)
The probability of one ace and one non-ace in either order is P(1)= 0.072+ 0.072= 0.144.

There are 52 cards of which 4 are aces. The probability the first card drawn is 4/52= 1/13. There at then 51 cards left of which 3 are aces. The probability the second card drawn is also an ace is 3/51. The probability both cards drawn are aces is [tex]P(2)= \frac{1}{13}\frac{3}{51}= P(2)= 0.0045