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    Aravinda's Avatar
    Aravinda Posts: 2, Reputation: 1
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    #1

    Apr 18, 2008, 03:54 AM
    probability of getting an ace from 50 cards if 2 cards are randomly removed from 52
    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's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #2

    Apr 18, 2008, 06:08 AM
    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's Avatar
    iamthetman Posts: 105, Reputation: 8
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    #3

    May 2, 2008, 08:44 AM
    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's Avatar
    Stratmando Posts: 11,188, Reputation: 508
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    #4

    May 3, 2008, 08:22 AM
    6.5 to 1 Odds?
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #5

    May 5, 2008, 08:21 AM
    Quote Originally Posted by 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.
    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.
    donna hall's Avatar
    donna hall Posts: 1, Reputation: 1
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    #6

    Mar 11, 2013, 05:27 PM
    two cards are drawn from 52 cards. What is the probability of drawing one ace and one face card (without replacement).
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #7

    Mar 12, 2013, 05:41 AM
    Donna: to determine the answer start with how many aces are there in the deck how many face cards? Now consider that there are two possible orders of drawing that will satisfy the condition - either scenario (1) you draw an ace first followed by a face card. Or scenario (2) you draw a face card first followed by an ace. Let's start with (1): the probability of drawing an ace followed by a face card is equal to (a) the number of aces divided by the number of cards in the deck times (b) the number of face cards divided by one less than the number of cards in the deck. Now do the same calculation for scenario (2) - what do you get? Finally the probabality of either scenarios (1) or (2) occurring is what? Post back with your answer and we'll check it for you.
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