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Math Help - probability!Please help

  1. #1
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    probability!Please help

    Eight tyres of different brands are ranked from 1 to 8 (best to worst) according to mileage performance. If four of these tyres are selected at random:

    a) Find the probability that the best tyre among those selected is actually ranked third among the original eight

    b) Find the probability that the worst tyre among those selected is actually ranked seventh among the original eight

    c) Find the probability that the best tyre and the worst tyre among these selected are actually ranked third and seventh respectively among the original eight. Are these two events independent

    d) Let random variable, R, represent the range of the original ranks in the four tyres selected. Thus in part (c), R = 7-3 = 4. determine the probability distribution for R and display in the form of a table

    e) Graph the probability mass function and cumulative distribution functions for R
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mika
    Eight tyres of different brands are ranked from 1 to 8 (best to worst) according to mileage performance. If four of these tyres are selected at random:

    a) Find the probability that the best tyre among those selected is actually ranked third among the original eight
    Suppose the first tyre chosen is the rank 3. Then the number of ways of
    choosing the remaining tyres so that they are all of greater rank is
    (as there are five tyres of lower rank than 3):

    <br />
5\times4\times3<br />

    but the rank three tyre could have been the 1st, 2nd, 3rd or 4th tyre
    drawn so there are four times as manny cases like this, and as we are interested
    only in combinations we must divide through by the number of permutations
    of four objects.

    So there are:

    <br />
N_1=\frac{4\times5\times4\times3}{4!}<br />

    different combinations with highest rank tyre being 3.

    To find the probability of this occuring we need to divide through by
    the number of combinations of tyres which is:

    <br />
N_2=\frac{8\times7\times6\times5}{4!}<br />

    So the required probability is:

    <br />
P(\mbox{highest rank=3)}=\frac{4\times5\times4\times3}{8\times 7 \times 6\times5}\approx 0.143<br />
    Last edited by CaptainBlack; March 15th 2006 at 10:00 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by mika
    Eight tyres of different brands are ranked from 1 to 8 (best to worst) according to mileage performance. If four of these tyres are selected at random:


    b) Find the probability that the worst tyre among those selected is actually ranked seventh among the original eight
    This is exactly like part a) except that if the first chosen tyre is of rank
    7, then there are six tyres of higher rank left to complete the selection

    So now:

    <br />
N_1=\frac{4\times6\times5\times4}{4!}<br />

    and we till have:

    <br />
N_2=\frac{8\times7\times6\times5}{4!}<br />

    so the required probability is:

    <br />
P(\mbox{lowest rank=7)}=\frac{4\times6\times5\times4}{8\times 7 \times 6\times5}\approx 0.286<br />

    RonL
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