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Math Help - Probability (cards)

  1. #1
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    Probability (cards)

    Just need confirmation for my answers ! Thank you!

    1. The chance that the top card in a well shuffed deck is a two or a nine is:
    8/52 = 15.4% (4 twos and 4 nines in deck)

    2. The chance that the eighth card from the top of the deck is a heart or a diamond is:
    26/52 = 50% (13 hearts & 13 diamonds in deck); assuming we don't know the first 7 cards; the position doesn't matter here.

    3. The chance that the top card in the deck is a two or a nine given that the eighth card from the top is a heart or a diamond is:
    Wouldn't it still be 15.4%? Unless they mention that the eighth card is a two or a nine, it wouldn't matter right?

    4. Are the events that the top card is a two or a nine and that the eighth card is a heart or a diamond independent?
    Yes (I'm not sure entirely on this one. An explanation would be great! Other options: No or Cannot Determined Without Knowing The Other Cards)

    Thanks a bunch!
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  2. #2
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    Re: Probability (cards)

    1 and 2 are fine
    for 3, i think the chances will be affected. Consider 2 cases:
    1:- the card we know of is neither 2 or 9
    2:- the card we know of is either 2 or 9 (of H or D)
    so
    P= (top card 2 or 9)*(case 1) + (top card 2 or 9)*(case 2)
    or (8/52)*(22/26)+(7/52)*(1/4)

    where 22/26 is chances that ~(2,9) was selected from D and H
    and 1/4 is chance that either of 2H,9H,2D or 9D was selected.

    about 4, no they are not independent because if the H or D is a 2 or 9, the chances of top card being 2 or 9 are affected.

    Hope it helps!
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  3. #3
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    Re: Probability (cards)

    For 3, there could only be one answer... so I'm not sure... I understand that there could be 2 scenarios: (1) neither 2 or 9 is in the 8th position (2) either 2 or 9 is in the 8th position. That being said, the information is limited? Could we assume that it's not 2 or 9 then? I'm confused ! Help please!

    Thanks you for the confirmation on 1, 2, 4 !!!
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  4. #4
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    Re: Probability (cards)

    Hello, Rikimaru!

    3. The probability that the top card in the deck is a 2 or a 9, given that the 8th card from the top is a \heartsuit or a \diamondsuit.

    Bayes' Theorem: . P(\text{top }2 \vee 9\,|\,\text{8th }\heartsuit \vee \diamondsuit) \;=\; \frac{P\big([\text{top }2 \vee 9]\:\wedge\:[\text{8th }\heartsuit \vee \diamondsuit]\big)}{P(\text{8th }\heartsuit\text{ or }\diamondsuit)} .[1]

    The numerator requires some work.
    The probability that the top card is a 2 or 9 depends whether the 8th card is the 2\heartsuit,\,9\heartsuit,\,2\diamondsuit,\text{ or }9\diamondsuit.

    [1] If the 8th card is not one of those 4 cards: . P = \frac{48}{52} \,=\,\frac{12}{13}
    . . .Then the top card is a 2 or 9 with probability \frac{8}{51}
    . . . P(\text{Case 1}) \:=\:\frac{12}{13}\cdot\frac{8}{51} \:=\:\frac{96}{663}

    [2] If the 8th card is one of those 4 cards: . P = \frac{4}{52} \,=\,\frac{1}{13}
    . . .Then the top card is a 2 or 9 with probability \frac{7}{51}
    . . . P(\text{Case 2}) \:=\:\frac{1}{13}\cdot\frac{7}{51} \:=\:\frac{7}{663}


    The numerator is: . \frac{96}{663} + \frac{7}{663} \:=\:\frac{103}{663}

    The denominator is: . {(\text{8th is }\heartsuit\text{ or }\diamondsuit) \:=\:\frac{26}{52}\:=\:\frac{1}{2}


    Therefore, [1] becomes: . \dfrac{\frac{103}{663}}{\frac{1}{2}} \;=\;\frac{206}{663}




    4. Are the events that the top card is a 2 or a 9 and that the 8th card is a \heartsuit or a \diamondsuit independent?

    No, as expained in part 3, the two events are dependent.

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  5. #5
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    Re: Probability (cards)

    So the final answer for 3 would be 206/663? What I'm confused here is that there are multiple scenarios... I think I might need to reread Bayes' Theorem since I recall seeing it in the book. Thanks for the clear up...!

    Um... I just resubmit the assignment with 206/663... and it seems like my previous answer: 15.38461% was right O_O!? Now I'm even more confused !
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  6. #6
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    Re: Probability (cards)

    Quote Originally Posted by Rikimaru View Post
    it seems like my previous answer: 15.38461% was right O_O!? Now I'm even more confused !
    Yes, your intuition was right. How could it matter??

    Quote Originally Posted by Soroban View Post
    [1] If the 8th card is not one of those 4 cards: . P = \frac{48}{52} \,=\,\frac{12}{13}

    [2] If the 8th card is one of those 4 cards: . P = \frac{4}{52} \,=\,\frac{1}{13}


    But don't forget that it's given that the 8th card is {\heartsuit\text{ or }\diamondsuit}, so we want

    [1] If the 8th card is not one of those 4 cards: . P = \frac{22}{26} \,=\,\frac{11}{13}

    [2] If the 8th card is one of those 4 cards: . P = \frac{4}{26} \,=\,\frac{2}{13}

    ... as pratique21 had it... although that poster seems to have somehow cancelled the second fraction down to a quarter, and also neglected to reduce the denominators under the 8 and 7 to 51 instead of 52.

    But for those slips, their calculation would have produced a fraction equal to 8/52.

    And, consequently, the two events are independent.


    Last edited by tom@ballooncalculus; July 18th 2012 at 01:55 PM.
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  7. #7
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    Re: Probability (cards)

    tom@ballooncalculus , thanks for pointing that out.
    However, could you explain how you came to conclude that the events are independent ?
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  8. #8
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    Re: Probability (cards)

    Quote Originally Posted by pratique21 View Post
    tom@ballooncalculus , thanks for pointing that out.
    However, could you explain how you came to conclude that the events are independent ?
    Quote Originally Posted by tom@ballooncalculus View Post
    But for those slips, their calculation would have produced a fraction equal to 8/52.
    And, I might have added, they were assuming...

    Quote Originally Posted by pratique21 View Post
    2 cases:

    (of H or D)
    ... i.e. that the 8th card is Hearts or Diamonds, which will happen in precisely half of the outcomes. So we have...

    P\big([\text{top }2 \vee 9]\:\wedge\:[\text{8th }\heartsuit \vee \diamondsuit]\big)\ =\ \frac{8}{52} \cdot \frac{1}{2}\ =\ P(\text{top }2 \vee 9)\cdot P(\text{8th }\heartsuit \vee \diamondsuit)

    ... satisfying...

    P(A \cap B) = P(A) \cdot P(B)

    Independence (probability theory) - Wikipedia, the free encyclopedia

    Just in case a picture helps...

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  9. #9
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    Re: Probability (cards)

    tom , kindly refer to the post by soroban, you will find that p(A and B) not equal to p(A)*p(B), so the events are dependent

    Logically, whether the eighth card that is drawn is a 2H,2D,9H or 9D is affecting the chances of top card being a 2 or a 9.
    So these events ca not possibly be dependent !
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  10. #10
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    Re: Probability (cards)

    Quote Originally Posted by pratique21 View Post
    Logically, whether the eighth card that is drawn is a 2H,2D,9H or 9D is affecting the chances of top card being a 2 or a 9.
    True. But that isn't the question. The question is...

    Quote Originally Posted by Rikimaru View Post
    3. The chance that the top card in the deck is a two or a nine given that the eighth card from the top is a heart or a diamond:
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  11. #11
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    Re: Probability (cards)

    For 3, there could only be one answer... so I'm not sure... I understand that there could be 2 scenarios:

    Loans and Grants
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