Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago

counting with combinations again?

five cards are dealt from a standard 52 card deck. find the probability of being dealt:

a) at least 1 black card ( isn't that1 - the probability of getting no black cards?)

b) a straight ( five cards in a sequence) example 7,8,9,10,jack, or 2,3,4,5,6 any suit

c) three of a kind

d) two pair

i have answers: a) .9747; b) .0035; c) .02113 d) .0475

thanks for helping out! just need a simple walk through to get the problems understood

Update:

I actually just asked the question exactly how my teacher asked it on the homework assignment. a straight could be any straight possible in the deck of cards beginning with ace-4, and up to 10-ace I am imagining.

Update 2:

kind is just rank, because there are 4 of each rank. 4 aces, 4 twos, 4- threes, 4-fours all the way to kings.

4 Answers

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  • 9 years ago
    Best Answer

    Your questions are vague.

    b) Will King, Ace, 2,3,4 form a sequence?Should the draw be in the sequential order?

    c) By kind do you mean suit?

    d) by pair what do you mean?

    a) P[at least one black] = 1 - P[none black]

    = 1 - P[all red] = 1 - 26C5/52C5 = 1 -0.02531= 0.97469

    b) Form an order as follows:

    Ace(1), 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace(1), 2, 3, 4, 5, 6,..........

    Here, the first card drawn can be any one of the 13 denominations in 52 ways. Once the first is fixed, the next should be one of the four following cards and the third should be one of the four succeeding cards and so on. thus the probability is (52x4x4x4x4)/(52x51x50x49x48) = 0.000042684.

    c) P[Three of same denomination and two are others] = 13(4C3*48C2)/52C5 = 0.02257

    d) P[two pairs and one other] = 13C2*4C2*4C2*44C1/52C5 = 0.00108

  • 9 years ago

    (a) It is 1-P(all red).

    P(all red)=(26/52)*(25/51)*(24/50)*(23/49)*(22/48)*(21/47) [because after you get your first red card, there is one less red card in the deck, and the deck has one less card, etc.]

    (b) For the numerator: there are 10 possible ranks of straights (five high, six high, ..., ace high).

    For each of the five cards in the hand, there are four choices for the suit: club, spade, heart, diamond).

    So the numerator is 10*(4^5). But you need to subtract 40 because there are 4*10 possible straight flushes. The denominator is just 52C5.

  • 4 years ago

    "the number cannot consist of all zero" implies that leading zeroes are permissible a. 10^4 - 1 = 9999 <------ b. 5*4 ways for 1st & last, 8*7 for others => 1120 <------ strictly speaking, codes, not 4digit #s, can have leading zeroes.

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