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Math Help - Probability question i dont understand

  1. #1
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    Angry Probability question i dont understand

    1. 2 card are chosen from a standard 52 card deck, what is the chances of getting the sum of them to be at least 4?
    aces = 1
    face cards = 10
    2. In a shuffled deck, what is the chances of ace of spades and ace of clubs to be adjacent?
    3.Someone flips 3 coins and rolls a fair die, what is the chances that the number of heads will equal the number showing on the die?
    Last edited by Sneaky; September 23rd 2010 at 11:52 AM.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    I'll proceed by reverse probability, that is, I'll find the probability of having the sum of the two cards less than 4.

    When does this happen? It's when you get 2 aces (sum = 2) or an ace and a two (sum = 3)

    No. of possibilities for 2 Aces = 4C2 = 6
    No. of possibilities for 1 Ace and a 2 = 4C1 x 4C1 = 16

    Total = 22.

    Now, total number of possibilities = 15C2 = 1326

    Hence, probability of having a sum of 4 or more (at least 4) is given by:

    P(sum \geq 4) = 1 - \dfrac{22}{1326}

    2. (EDIT: Removed as question was misread, sorry)

    I hope it helps!
    Last edited by Unknown008; September 23rd 2010 at 12:45 PM.
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  3. #3
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    what about number 3?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    When is the event true?

    When you have one head, and you get a 1 on the dice. (a)
    Or, when you have 2 heads and 2 on the dice. (b)
    Or, when you have 3 heads and 3 on the dice. (c)

    Probability for (a) = 3(\frac{1}{2})^3 \times \frac16

    1/2 for head, 1/2 for tail and 1/2 for tail, in any order, that is (1/2 x 1/2 x 1/2) x 3. 1 out of 6 possibilities gives a 1 on the dice.

    Probability for (b) = 3(\frac{1}{2})^3 \times \frac16

    1/2 for head, 1/2 for head and 1/2 for tail, in any order, that is (1/2 x 1/2 x 1/2) x 3. 1 out of 6 possibilities gives a 2 on the dice.

    Probability for (c) = 3(\frac{1}{2})^3 \times \frac16

    I think you might guess why by now.

    Total probability is the sum.

    EDIT: For (c), there is not multiplying by 3.
    Last edited by Unknown008; September 23rd 2010 at 12:35 PM.
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  5. #5
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    for your answer to question 2, why did u put 1/52x1/51, the 1/52 part is like saying that the ace of spades has to be in a specific spot right? i did
    1/51 + 1/51 = 2/51, because don't you only need to count the chance of the club one being beside the spade one regardless of where the spade one is?

    And what about number 3? I don't understand it at all..
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Hm... that's why I then added 1/52 x 1/51.

    Take this scenario:
    First is ace of spade, second is ace of clubs. Probability = 1/52 x 1/51

    okay

    Second case,
    First is an ace of clubs, second is an ace of spades. Probability here is = 1/52 x 1/51
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  7. #7
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    for your answer to number 3, i see why u did 1/2 ^ 3, because that is the chance of getting HHH or HHT or HTT respectively, but i dont get why u multiplyed it by 3 after...

    the number of combinations for the coins is
    HHH,HHT,HTH,HTT
    TTT,TTH,THT,THH

    only one of which has 3 H's, so 3 heads is 1/8 chance
    so for your answer for (c)
    shouldnt it be
    1/2^3 * 1 * 1/6
    ??
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  8. #8
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    Hi guys,

    concerning the second question, I think 2 / 52 is correct. Try it this way: You've got 52 slots for the cards and 52! possibilites to fill those slots. Let's say the ace of spades is in the first slot, the ace of clubs in the second. This leaves 50! possibilities to fill the remaining slots. But we're not restricted to the first two slots: there are 51 further possibilites to choose two adjecent slots. Furthermore, as Unknown008 already pointed out, you can switch the ace of spades and the ace of clubs and they are still neighbours; this gives rise to a factor of 2! or 2. So in the end we have 2 * 51 * 50! / 52! = 2 / 52 confirming Unknown008's result.

    I hope this helped!

    Greetings,
    needsomemathhelp
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  9. #9
    MHF Contributor Unknown008's Avatar
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    Yes, sorry, good catch. I was doing too many things at once, sorry.

    Yes, for 1 head, you have HTT, THT and TTH
    For 2 heads, you have HHT, HTH and THH
    For three heads, you have HHH only.
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  10. #10
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    Ok thanks for telling me

    btw i am still stuck on

    2. In a shuffled deck, what is the chances of ace of spades and ace of clubs to be adjacent?

    I dont get if the answer is 2/52 or 1/1326 .....
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  11. #11
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    @Sneaky,

    you're correct, the probability for having 3 heads is 1 / 8 only. So for your problem No. 3 the answer is 7/ 48.
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  12. #12
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    but what about
    2. In a shuffled deck, what is the chances of ace of spades and ace of clubs to be adjacent?

    I dont get if the answer is 2/52 or 1/1326 .....

    I also dont understand this question.
    Suppose we roll 3 fair die, what is the chances that 2 match and one is different.
    i put for this
    there are 216 outcomes
    chance of getting a number on die1 = 6/6
    chance of getting the same number as on die1 in die2 = 1/6
    chance of getting a different number as die1 in die3 = 5/6
    6/6 * 1/6 * 5/6 = 5/36
    Is this right?
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  13. #13
    MHF Contributor Unknown008's Avatar
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    Darn, sorry for number 2, I took the question as being if two cards are chosen at random...

    Okay, let treat is another way.


    There are 52 cards. Now, let's group the ace of clubs and the ace of spades. This makes 51 items in all. So, there are 51! combinations and we permute the two aces in question between them, making 2 more possible combinations.

    This gives 51! * 2 combinations.

    There is a total of 52! combinations.

    The probability thus becomes \dfrac{51! \times 2}{52!} = \dfrac{1}{26}
    Last edited by Unknown008; September 23rd 2010 at 12:51 PM. Reason: replaced /frac by /dfrac
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  14. #14
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    @Unknown008,

    I don't see why there should be 51! combinations: With two cards chosen, only 50 remain and hence 50! combinations of the non-ace-of-spades/non-ace-of-clubs cards. But we agree on
    2 \frac{51 \cdot 50!}{52!}. You just put the additional factor of 51 in the factorial...

    2 = number of permutations of the pair of aces
    51 = number of positions of the pair of aces in the deck
    50! = number of combinations of the remaining cards
    52! = number of combinations of the complete deck
    Last edited by needsomemathhelp; September 23rd 2010 at 12:56 PM.
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  15. #15
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    that was what unknown008 got with his initial answer, then changed his method
    But i am not understanding his new method.
    The way i see it is:

    in the deck there is an ace of spades.
    now the possible slots left is 51
    of the 51, either 1/51 or the other 1/51 can be filled with ace of clubs to make it true, so then there is a 2/51 chance of this being true.
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