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Math Help - probability of choosing two word

  1. #1
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    probability of choosing two word

    Hi

    Can anybody tell wether i am right or not.

    Q. Two WORDS are chosen from word "SUMMER", what is probability that at least M is there?

    Total Nos of chosen word are : 6C2=15
    SU SM SM SE SR
    UM UM UE UR
    MM ME MR
    ME MR
    ER

    SO P(at least one is M)=9/15

    Thanks in Advance...
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  2. #2
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    Ok, Yes your answer is right.

    Something seemed odd so I used to the hypergeometric distribution formula to make sure. 9/15 is the correct answer

    \frac{{{2}\choose{1}}{{4}\choose{1}}}{{{6}\choose{  2}}} +\frac{{{2}\choose{2}}{{4}\choose{0}}}{{{6}\choose  {2}}}=\frac{9}{15}
    Last edited by downthesun01; October 5th 2010 at 02:06 AM.
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  3. #3
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    Thank you...
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  4. #4
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    Hello, amb03!

    \text{Can anybody tell whether i am right or not.?}

    \text{A two-letter "word" is chosen from word SUMMER.}
    \text{What is probability that at least one M is chosen?}


    \text{Total number of chosen words is: }\;_6C_2\:=\:15

    . . \begin{array}{c}<br />
\text{SU SM SM SE SR} \\<br />
\text{UM UM UE UR} \\<br />
\text{MM ME MR} \\<br />
\text{ME MR} \\<br />
\text{ER} \end{array} .Good!


    \text{So: }\;P(\text{at least one M}) \:=\:\dfrac{9}{15} . Correct!

    I solved it from "the other end" . . .


    How many words have no M's?


    There are 2 M's and 4 Others.

    We want 2 of the Others.
    . . There are: . _4C_2 \:=\:6 words with no M's.

    Hence, there are: . 15 - 6 \:=\:9 words with at least one M.

    Therefore: . P(\text{at least one M}) \:=\:\dfrac{9}{15}
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