# Math Help - Conditional Probability

1. ## Conditional Probability

I have the following probability problem I am trying to figure out how to go about. It seems to be a conditional probability but it seems more intuitive than that.

Suppose that a hand of 5 cards is selected at random, without replacement, from a standard deck of 52 playing cards.

(a) If it is known that the hands contains the ace of spades, what is the probability that the hand contains exactly two aces?

(b)If it is known that the hand contains at least one ace, what is the probability the hand contains exactly two aces?

I am on part (a)

Well If I just look at the first probability, which is, the 5-card hand contains an ace of spade I see it as this

1 ace of spades out of 4 total aces
4 non-aces out of 48 remaining non-ace cards
then

Pr(the hand contains the ace of spades)=
$(^{4}_1)*(^{48}_4)$

correct?

Then the hand containing two aces given you know the hand contains the ace of spades would be.
$(^{4}_2)*(^{48}_3)$

If it seems that the condition of knowing the hand contains an ace of spades does not really affect the probability the hand could contain two aces.
Intutively for me this makes some sense but I feel like I leaving something out. Also, it seems like part (b) would be exact the same as part (a)

All help is appreciated

2. ## Re: Conditional Probability

Originally Posted by IHuntMath
Suppose that a hand of 5 cards is selected at random, without replacement, from a standard deck of 52 playing cards.

(a) If it is known that the hands contains the ace of spades, what is the probability that the hand contains exactly two aces?
There $\binom{51}{4}$ five card hands that contain the ace of spades.

There $3\cdot\binom{48}{3}$ five card hands that contain the ace of spades and exactly one other ace. Divide.

3. ## Re: Conditional Probability

Originally Posted by Plato
There $\binom{51}{4}$ five card hands that contain the ace of spades.

There $3\cdot\binom{48}{3}$ five card hands that contain the ace of spades and exactly one other ace. Divide.
Thanks for the quick reply, let them try to decipher into my own terms.

Discounting the fact there is already one known card, that is, ace of spades. Leave the remaining 51 cards choosing 4 cards to fulfill a 5 card hand since you already have one, the ace of spades. Is this the reasoning for $\binom{51}{4}$

Where does the 3 come from in $3\cdot\binom{48}{3}$
48 remaining cards choose 3, time 3?

Sorry, to be a bit dumb. Here is what I came up with after I posted this. Not sure if it is correct but here is my reasoning
I don't see how knowing you have an ace in your 5-card hand affects the outcome of a 2 ace 5-card hand. These seem independent to me.

The probability you have an ace in your 5-card hand

1 ace out of 4 possible aces
4 remaining cards in the hand out of 48 non-ace cards

$\frac{\binom{4}{1} \cdot \binom{48}{4}}{\binom{52}{5}}$

where $\binom{52}{5}$ is the number of ways you can make a 5-card hand.

Then for a 2 ace 5-card hand

$\frac{\binom{4}{2} \cdot \binom{48}{3}}{\binom{52}{5}}$