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Math Help - some easier probability

  1. #1
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    some easier probability

    1. I have a card deck with 52 cards and I have to pick 3 cards and what are the probability of these events and also a opposite events
    p(A)- 1 black and 2 red
    p(B) -3 reds
    p(C) - 1 club, 1 diamond and 1 heart

    2. There are 10 white and 22 black balls. How big is the probability if I pick 2 white and 2 black balls?

    I out of my mind and I doesnt understand how to solve them. And excuse me for my english.
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  2. #2
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    Hello averera

    Welcome to Math Help Forum!
    Quote Originally Posted by averera View Post
    1. I have a card deck with 52 cards and I have to pick 3 cards and what are the probability of these events and also a opposite events
    p(A)- 1 black and 2 red
    p(B) -3 reds
    p(C) - 1 club, 1 diamond and 1 heart

    2. There are 10 white and 22 black balls. How big is the probability if I pick 2 white and 2 black balls?

    I out of my mind and I doesnt understand how to solve them. And excuse me for my english.
    Here's how you do 1A:

    The total number of ways of choosing 3 cards from 52 is
    \binom{52}{3}=\frac{52!}{3!49!}=\frac{52.51.50}{3.  2}=26.17.50
    The number of ways of choosing 1 black from 26 is 26; and the number of ways of choosing 2 red from 26 is \binom{26}{2}. So the number of ways of choosing 1 black and 2 red is
    26\times\binom{26}{2}=\frac{26\times26!}{2!24!}=\f  rac{26.26.25}{2}=26.13.25
    Therefore the probability that this occurs when 3 cards are chosen from 52 is
    \frac{26.13.25}{26.17.50}=\frac{13}{34}\approx 0.382
    I'm not quite sure what you mean by the 'opposite events'. Do you mean the probability that this doesn't happen? If so, then it's 1 - 0.382 = 0.618.

    Do 1B and 1C in a similar way.

    Start B by saying "The number of ways of choosing 3 reds from 26 is \binom{26}{3} ..."

    And start C by saying "The number of ways of choosing 1 club from 13 is 13...

    Can you complete these now?

    For question 2:

    The number of ways of choosing 4 balls from 32 is \binom{32}{4} = ... ?
    (a)

    The number of ways of choosing 2 white from 10 is ... ?

    The number of ways of choosing 2 black from 12 is ... ?

    So the number of ways of doing each of these things, one after the other is ... ?
    (b) (Hint: Multiply these two answers together.)

    Therefore the probability that this occurs is ... ? (Hint: divide the second answer
    (b) by the first, (a).)

    Can you do this one now?

    Grandad

    PS Here are the answers.
    Spoiler:
    1B. \frac{2}{17}\approx 0.118

    1C. \frac{169}{1700}\approx 0.099

    2. 0.289
    asdf
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