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Thread: Question about probability: Peta deals with a hand of 10 cards...

  1. #1
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    Question Question about probability: Peta deals with a hand of 10 cards...

    The question that I have problem is:

    Peta deals with a hand of 10 cards from a well-shuffled pack of ordinary playing cards.
    Show that the probability that she deals exactly 5 spades is less than 5%.


    There are 52 cards all together, out of which 13 are spades so the probability of getting a spade would be 1/4, which is 0.25.

    So the probability of dealing exactly 5 spades out of 10 would be, using the formula for binomial distribution, (10C5)*(0.25^5)*(0.75^5) = 0.0584 (3 s.f.), which is not less than 0.05.

    However, the answer section for this question in the textbook says that the answer is:
    0.0468... = 4.7 % (2 s.f.) < 5 %.

    But I have no idea how they got this figure because the book does not show any working, and I do not understand why I cannot use binomial for this question.

    I would much appreciate it if someone can help me to understand why I cannot use binomial here and why the probability is about 4.7 % as above.

    Or am I correct and the textbook is wrong?

    Thank you.

    (P.S. This question is under the chapter "Probability" in the textbook, which is placed BEFORE the chapter for "Binomial Distribution", therefore, there is supposed to be a way to work it out without using binomial formula)

  2. #2
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    The reason why you can't use the binomial formula here is that it only works if the probability of each event is constant. That's not the case here. You've correctly identified that the probability of the first card drawn being a spade is 13/52 = 1/4. But after that first spade is drawn, there's only 12 spades left in the deck, out of 51 cards. Because of the commutative nature of multiplication, you can safely assume that the first five cards drawn are spades and the other five cards are non-spades, without any loss of generality. So, after drawing five spades, how many cards are left in the deck? Of those, how many are non-spades?

  3. #3
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    Smile Thank you for your help.

    Thank you very much for your help.

    Now I see how to work it out properly.
    As you say, the probability of success is not constant, so we need to change the probability every time you deal a card.
    So, the probability of getting a spade is 13/52, but after that, the total number of cards and the number of spades get one less each time, so the probability of getting a spade five times in a row would be (13/52)*(12/51)*(11/50)*(10/49)*(9/48).
    Then, after that, a failure must follow 5 times in a row, so similarly the probability of NOT getting a spade (i.e. getting a non-spade card) 5 times in a row would be:
    (39/47)*(38/46)*(37/45)*(36/44)*(35/43).
    So, the probability of getting a spade 5 times in a row and then getting a non-spade 5 times in a row would be:
    (13/52)*(12/51)*(11/50)*(10/49)*(9/48)
    *
    (39/47)*(38/46)*(37/45)*(36/44)*(35/43)
    However, there are (10C5) different ways of getting spades or non-spades, for example,

    1 2 3 4 5 6 7 8 9 10
    S S S S S F F F F F
    F S F F S S F S F S
    ..........

    So it has to be multiplied by 10C5, correct?

    Therefore, a correct working and answer would be:
    (13/52)*(12/51)*(11/50)*(10/49)*(9/48)*(39/47)*(38/46)*(37/45)*(36/44)*(35/43)
    *(10C5) = 0.0468393... is approximately 4.7% (2 s.f.), which is less than 5%.

    Thank you very much again for your kind help.

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