probability

[luvugodm]

Find the probability of drawing a three card hand that includes two aces from a deck of 52 cards. Write answer as a fraction.

 

[pka]

\(\displaystyle \L\\\frac{C(4,2)C(50,1)}{c(52,3)}\)

That is an over-counting.
There are \(\displaystyle {4 \choose 2}\(48\) + {4 \choose 3}\) ways to choose a 3-card hand with at least two aces.

There are \(\displaystyle {4 \choose 2}\(48\)\) ways to choose a 3-card hand with exactly two aces.

 

[galactus]

Hey pka. I fail to see how that was wrong, but I will take your word for it.

It's pretty close though.

(4/52)(3/51)*3=3/221

3/221=.01357466...

72/5525=.01303167....

 

[pka]

If we pick two aces and then one more card from the fifty left, we may pick one of the two aces left. Therefore, we would have counted selecting any three aces twice.

There are \(\displaystyle {4 \choose 2}\(48\) + {4 \choose 3} = {292}\) ways to choose a 3-card hand with at least two aces.

There are \(\displaystyle {4 \choose 2}\(48\) = {288}\) ways to choose a 3-card hand with exactly two aces.

There are \(\displaystyle {{52} \choose 3} = {22100}\) ways to choose any 3-card hand .

 

[galactus]

Of course, pka. I feel DUH.

Doing it my more complicated way, I think we would have:

3(4/52)(3/51)-6(1/50)(4/52)(3/51)=72/5525

This should account for the subtraction of the other Ace scenario.