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Probability problem Watch

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    A manufacturer has six distinct motors in stock, two of which came from a particular supplier. The motors must be divided among two production lines, with three motors going to each line. If the assignment of motors to lines is random, find the probability that both motors from the particular supplier are assigned to the first line.

    I'd appreciate an explanation on how to work out the answer to this one.
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    Note that all outcomes, i.e. all arrangements of the six motors, are equally likely. Therefore the probability you seek is the number of outcomes in which the two from the particular supplier end up on the first line divided by the number of outcomes


    (Original post by The OP)
    A manufacturer has six distinct motors in stock, two of which came from a particular supplier. The motors must be divided among two production lines, with three motors going to each line. If the assignment of motors to lines is random, find the probability that both motors from the particular supplier are assigned to the first line.I'd appreciate an explanation on how to work out the answer to this one.
    edit: well, better not to think of all the arrangements, that'd be a lot, but rather how they can be divided amongst the lines
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    nice question
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    (Original post by The OP)
    A manufacturer has six distinct motors in stock, two of which came from a particular supplier. The motors must be divided among two production lines, with three motors going to each line. If the assignment of motors to lines is random, find the probability that both motors from the particular supplier are assigned to the first line.

    I'd appreciate an explanation on how to work out the answer to this one.
    Find the total number of ways to distribute the six distinct motors over two groups which have a size of 3:
    6!/(3!3!) = 20
    Assume that the condition is met, we then have 4 motors left but now the first line is 2/3rds full, the number of ways this can happen is:
    4!/(1!3!) = 4

    Hence the probability is 4/20 = 1/5

    I may be wrong, its early.
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    Or calculate how you could have neither of those two in the other line.
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    (Original post by SCalver)
    Find the total number of ways to distribute the six distinct motors over two groups which have a size of 3:
    6!/(3!3!) = 20
    Assume that the condition is met, we then have 4 motors left but now the first line is 2/3rds full, the number of ways this can happen is:
    4!/(1!3!) = 4

    Hence the probability is 4/20 = 1/5

    I may be wrong, its early.
    That's what I got
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    the problem is beautifully easy

    say AABBBB and we require the two As to be together

    we pick the 1st P(B) =4/6
    we pick the 2nd P(B) =3/5
    we pick the 3rd P(B) =2/4

    that means that the 2 As are now together

    hence required probability is 1/5 (by multiplying the above fractions)
 
 
 
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