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Math Help - couple of probability questions...

  1. #1
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    couple of probability questions...

    There are eight pairs of shoes with different styles. Each customer can pick 2 shoes out of the bucket.

    (a) What is the probability that at least one customer obtains two shoes from the same pair if 2 customers pick in succession(no put backs) ?

    I think i need to find the probability all correct ways:
    1st and 2nd succeeds:
    1/15 +1/13
    1st person fails, 2nd person succeeds:
    14/15 *12/14*1/13
    1st person succeeds, 2nd fails:
    1/15*12/13
    total ~= .262


    (b) How much would an individual customer improve his chances of getting at least one complete pair by taking four shoes?
    Chance of success with 2 picks = 1/15
    with 2 additional pick customer can...
    1st 2 picks different pair, 3/4 pick matches correct pair: 14/15*2/14
    1st 2 picks different pair, 3/4 pick is a new pair: 14/15 * 12/14*1/13
    1st 2 picks same pair, 3/4 pick another pair: 1/15*1/13
    result = .1991 for 4 picks
    for 2 picks 1/15=.067 so .1991-.067 = .1321 more chance in improvement?


    The color of a person's eyes is determined by a single pair of genes.
    If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed
    genes, then the person will have brown eyes; and if one of them is a blue-eyed gene and the other is
    a brown-eyed gene, then the person will have brown eyes. A newborn child independently receives
    one eye gene from each of its parents and the gene it receives from a parent is equally likely to be
    either of the two eye genes of that parent. Suppose that John and both of his parents have brown
    eyes, but John's sister has blue eyes.

    (a) What is the probability that John possesses a blue-eyed gene?
    Now, suppose that John is also married and his wife has blue eyes.
    (b) What is the probability that their rst child will have blue eyes?
    (c) If their rst child has brown eyes, what is the probability that their next child will also have
    brown eyes?

    a. Well assuming John's parents both have BrBl gene then i put it on a table and there are 4 possible combinations - BrBr,BrBl,BlBr,BlBl.. so i said .75

    b. since his wife has BlBl she doesnt come into the picture so it is john again - since we know he has a one of the 4 combinations then i broke it down do what chances for a Blue eye gene. He can have .5 chance for BrBl and .25 chance for BlBl. For BlBl it is just .25 while the other two BrBl/BlBr it is .5*.5=.25+.25 = .5

    c. Since the eye gene is independent it is irrelevant the first child has brown eyes - i basically did like what i did for b except BrBl,BlBr and BrBr are used so i get .5 again
    Please help! Thank you!
    Last edited by shaselai; January 22nd 2009 at 07:58 AM.
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  2. #2
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    Quote Originally Posted by shaselai View Post
    I am confused about a couple of questions.. hope you guys can help...
    Failure rates of G,F,B
    p(G) = .1
    p(F) = .1 p(F) = .5 if G or B fails

    p(B) = 0 p(B) = .5 if G fails Mr F says: This data does not make sense. Please re-check.

    a. what is the probability that all three fail? I think it is just .1*.5*.5 = .025

    b Given that more than one failed, determine the conditional probabilities that:
    1. G did not fail.
    Since G is independent is it just 1-.1 = .9?
    2. B failed.
    Not sure...
    3. Both the G and F failed.
    Not sure either.. B depends on G so B could fail too?
    [snip]
    ..
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