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Combinations and their application to probability

  1. Jul 8, 2006 #1
    I am doing some work on high school (Australia) level combinatorics. So I've been taught nCr, nPr, and of course factorial (!). Now I need to apply the combinatorics to probability.

    There a two questions which I am stuck on.

    I calcualted that the sample space of all combinations was [itex]2^8-1[/itex] since there has to be at least one ingredient for there to be a samwhich.

    For part (a):
    [tex]Pr(samwhich contains ham)=\frac {n(Ham)}{2^8-1}[/tex], where n(ham) is the number of combinations including ham.

    The answer indicates that [itex]n(ham) = 2^7[/itex], but I don't know why it is [itex]2^7[/itex].

    Part (b) is:
    [tex]\frac {^8 C_3}{2^8-1}[/tex]

    Part (c) I worked out to be (which is right according to the book):
    [tex]\frac {\sum_{n=3}^8 ^8 C_n}{2^8-1}[/tex]

    Another problem question for me is:
    From that information I caculated that [itex]n(\epsilon) = ^18 P_3[/itex]. From that I caculated for (a) that the probaility of all red is:

    [tex]Pr(all red) = \frac {5*4*3}{^18 P_3} = \frac {60}{4896} = \frac {5}{408}[/tex]

    But the answer in the book is 5/204, so I'm off by a factor of 2, and I don't know why.

    I calcualted part be to be:

    [tex]Pr(all different colors) = \frac {7}{136}[/tex]

    But the book says the answer is:

    [tex]Pr(all different colors) = \frac {35}{136}[/tex]
     
    Last edited: Jul 8, 2006
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  3. Jul 8, 2006 #2

    arildno

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    Well, ask yourself:
    How many combinations do NOT include ham?
     
  4. Jul 8, 2006 #3
    Ah it would be:

    [tex]\sum_{n=1}^7 ^7 C_n = 2^7[/tex]

    Thanks for that.

    I would also be extremely appreciative if anyone could help me on the second question in my original post.
     
  5. Jul 8, 2006 #4

    0rthodontist

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    Actually, not quite for those sandwiches that do not include ham--remember that you're not counting sandwiches with 0 ingredients. That is correct for those that do include ham.

    In your second question, the numerator should not be 5 * 4 * 3. First you choose from among 6 red marbles, then you choose from among 5 red marbles, then you choose from among 4 red marbles.

    In the other part of it, you forgot to consider permutations. You've given only the probability of choosing White, Red, Blue. But also Red, White, Blue, or Blue, Red, White, or the other permutations of the three colors should be counted.
     
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