Another question concerning playing cards...?

You pick 8 cards from a deck. What is the probability that among the 8 will be...3 faces, 2 aces and 3 other number cards? Thanks

Update:

each card is replaced after looking at it

6 Answers

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  • Dr D
    Lv 7
    1 decade ago
    Best Answer

    If face cards are J,Q,K, then there are 12 of them.

    There are 4 aces and 36 others.

    There are 4C2 ways of picking 2 aces, 12C3 ways of picking 3 faces, and 36C3 ways of picking 3 others.

    Total = 4C2 * 12C3 * 36C3

    = 9,424,800

  • 1 decade ago

    There are 12 face cards, 4 aces and 36 number cards in a deck.

    Imagine the picking of the cards one by one.

    The probability of choosing the first face card is 12/52.

    The probability of choosing the second face card is 11/51.

    The probability of choosing the third face card is 10/50.

    The probability of choosing an ace after that is 4/49.

    The probability of choosing another ace is 3/48.

    The probability of choosing a face after that is 36/47.

    The probability of choosing another face is 35/46.

    The probability of choosing another face is 34/45.

    Therefore, the probability of choosing the 8 cards is

    = 12/52 X 11/51 X 10/50 X 4/49 X 3/48 X 36/47 X 35/46 X34/45

    = 11/491855

  • 1 decade ago

    You get the answer from the multivariate hypergeometric distribution.

    where ∑m = N and ∑k = n

    f(m1,m2,...,mj,k1,k2,...ki) = ∏(maCka) / (NCn)

    So in this case we have N=52 cards, n=8 cards being drawn, m1=12 face cards to select from, m2=4 aces, m3=52-12- 4=36 and k1=3 faces to select, k2=2 aces, and k3=8-3-2=3

    So f(m1,m2,m3,k1,k2,k3) = 12C3*4C2*36C3 / 52C8

    12C3 = 12!/[(12-3)!3!] = 12*11*10/(3*2) = 220

    4C2 = 4!/[(4-2)!2!] = 4*3/2 = 6

    36C3 = 36!/[(36-3)!3!] = 36*35*34/(3*2) = 7,140

    52C8 = 52!/[(52-8)!8!] = 752,538,150

    f = 220*6*7,140/752,538,150 = 9,424,800/752,538,150

    f = 0.012524 = 1.2524% (roughly 1 in 80)

  • cidyah
    Lv 7
    1 decade ago

    nCr = n! / r! (n-r)! , number of ways of choosing r things out of n things.

    8 cards may be chosen in 52C8 ways.

    52C8=52! / 8! 44! -- (1)

    There are 12 face cards. 3 can be chosen in 12C3 ways. 2 aces can be chosen in 4C2 ways. 3 number cards may be drawn in 36C3 ways.

    Probability= (12C3)(4C2)(36C3) / (52C8)

    Simplify the above expression.

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  • 1 decade ago

    there are 12 face cards

    4 aces and 36 regular cards

    there you go...you have all the info to figure it out.

  • 1 decade ago

    12f + 4a + 36n = 52 in all

    P(3f,2a,3n)

    = (12/52)^3 * (4/52)^2 * (36/52)^3

    = 1289945088 / 53459728531456

    = 0.002413%

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