Re: Conditional Probability

Quote:

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 $\displaystyle \binom{51}{4}$ five card hands that contain the ace of spades.

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

Re: Conditional Probability

Quote:

Originally Posted by

**Plato** There $\displaystyle \binom{51}{4}$ five card hands that contain the ace of spades.

There $\displaystyle 3\cdot\binom{48}{3}$ 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 $\displaystyle \binom{51}{4}$

Where does the 3 come from in $\displaystyle 3\cdot\binom{48}{3}$

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

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

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

Then for a 2 ace 5-card hand

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