From a pack of 52 cards 2 cards are randomly removed. Then what is the probability of getting an ace from remaining 50 cards?

galactus

There are 3 cases to consider. If two aces are drawn from the 2 selected, if 1 ace is drawn from the 2 selected, or if no aces are drawn from the two selected.

Find the probability for each case and add them up.

iamthetman

Actually this is a case for conditional probability.

Here's how you solve conditional probabilities:
P(A) = P(A|B)xP(B)

P(getting ace from 50 cards) = P(getting ace from 50 cards|4 aces left)xP(4 aces left) + P(getting ace from 50 cards|3 aces left)xP(3 aces left) + P(getting aces from 50 cards|2 aces left)xP(2 aces left)

Stratmando

6.5 to 1 Odds?

ebaines

galactus - I think you answered the wrong question - your method would help find the probability that one of the 2 discarded cards is an ace. But I think what the OP was asking is actually this:

Take a deck of 52 cards. Discard 2, leaving 50. Now draw one card from the remaining 50 - what is the probability that this card is an ace?

IATM has the mathematically correct answer. But this problem is one of those that really doesn't require such high-powered mathematics. It's a bit like swatting a fly with a sledge hammer- it works, but something a little simpler may be easier. Here restating the problem can make it a lot easier to understand - consider that the OP's question is identical to the following:

Start with a deck of 52 cards and deal the cards one by one - what is the probability that the third card drawn is an ace?

Now the answer should be pretty obvious.