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Homework Help: Statistics Question

  1. Jul 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Three cards are dealt from a well-shuffled deck.

    (a) Find the chance that all of the cards are diamonds.
    (b) Find the chance that none of the cards are diamonds.
    (b) Find the chance that the cards are not all diamonds.

    2. Relevant equations

    Not sure ...

    3. The attempt at a solution

    There are 52 cards in a deck. The chance of one being a diamond is 13/52. If you know that one card you were dealt is a diamond, then the chance of another being a diamond is 12/51, since there is 1 less card and 1 less diamond. And if you that two cards are diamonds, then the chance of a third being a diamond is 11/50. The chance of 3 cards being diamonds is therefore (13/52)*(12/51)*(11/50). Part (b) can be solved similarly. This time the chance is 39/52, then 38/51, then 37/50, and the chance of the three happening is (39/52)*(38/51)*(37/51). Part (c) is simply the opposite of part (a).

    Am I right?
  2. jcsd
  3. Jul 24, 2010 #2
    It sounds like you have this all worked out to me. You have the probabilities for each stage correct, and multiplying the three together should lower the probability to show the probability of the two states, and then three. It looks good to me.

    I came up with 11/850 for part(a) using that method.

    I'm not sure that I entirely understand part(c).
    "Find the chance that the cards are not all diamonds."

    Sounds like this one is a little different. If the first card is a diamond, so it the second, but the third is not, it would still satisfy this condition right? I think this one is asking "What are the chances that none of the cards will be a diamond, and all of the cards will not be diamonds"

    I don't think it's the opposite of part(a) as you mentioned.
    Last edited: Jul 24, 2010
  4. Jul 24, 2010 #3
    I'm just confused because this chapter was supposed to be about adding things that are mutually exclusive.
  5. Jul 24, 2010 #4
    I don't know what the chapter was supposed to be about, but I don't really see any other solution.

    Not all the cards are diamonds...

    So lets think of this in three cases.
    In case 1, let's say the card cannot be a diamond
    (3/4) right?

    Let's also assume that because cases 1 and 2 will fulfill the "not all be diamonds" requirement, case three can be anything.

    so what would case 2 have to be?
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