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Math Help - [SOLVED] probability word problem help

  1. #1
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    [SOLVED] probability word problem help

    A boy has 6 red and 7 blue blocks.
    (i) in how many ways can he choose randomly 4 blocks.
    i got 715 ways

    (ii)what is the probablity that he will chose for red blocks.
    i got 15/715 but i think is wrong.

    (iii) what is the probability that the choice includes at least one blue block

    thanks
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  2. #2
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    i) P(X=4) = \binom{13}{4}

    ii) P(X=4 red) = \binom{6}{4} / \binom{13}{4}

    iii) P(X=1 blue) = \binom{7}{1} \binom{6}{3} / \binom{13}{4}

    P(X=2 blue) = \binom{7}{2} \binom{6}{2} / \binom{13}{4}

    P(X=3 blue) = \binom{7}{3} \binom{6}{1} / \binom{13}{4}

    P(X=4 blue) = \binom{7}{4} / \binom{13}{4}

    Answer is P(1)+P(2)+P(3)+P(4)
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  3. #3
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    thanks.
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  4. #4
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    Quote Originally Posted by llkkjj24 View Post
    A boy has 6 red and 7 blue blocks.
    (i) in how many ways can he choose randomly 4 blocks.
    i got 715 ways

    (ii)what is the probablity that he will choose four red blocks.
    i got 15/715 but i think is wrong. why ?

    (iii) what is the probability that the choice includes at least one blue block

    thanks
    If the blue blocks all look the same and the red blocks all look the same, then

    (i)

    \binom{13}{4}=715

    Or....

    \color{red}\binom{6}{4}=15=number\ of\ ways\ to\ choose\ 4\ red\ blocks

    \color{blue}\binom{7}{4}=35=number\ of\ ways\ to\ choose\ 4\ blue

    \color{red}\binom{6}{3}\color{blue}\binom{7}{1}=14  0=\color{red}3\ red\color{blue}\ with\ 1\ blue

    \color{red}\binom{6}{2}\color{blue}\binom{7}{2}=31  5=\color{red}2\ reds\color{blue}\ with\ 2\ blues

    \color{red}\binom{6}{1}\color{blue}\binom{7}{3}=21  0=\color{red}1\ red\color{blue}\ with\ 3\ blues

    total=715

    (ii)

    Why do you think your answer is wrong??

    It was ok to be suspicious, however \binom{13}{4}=715

    You get the same answer by summing the individual cases.


    (iii)

    the purpose of the question is for you to discover that there is a fast way to calculate (iii)
    At least 1 is blue is equivalent to "the 4 are not all red".

    Since all probabilities sum to 1,
    the answer is

    1-\color{red}P(all\ 4\ are\ red)\color{black}=1-\frac{\color{red}\binom{6}{4}\color{black}}{\binom  {13}{4}}=1-\frac{\color{red}15}{75}
    Last edited by Archie Meade; May 2nd 2010 at 04:06 AM. Reason: small typo
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    If the blue blocks all look the same and the red blocks all look the same, then

    (i)

    \binom{13}{4}=715

    Or....

    \color{red}\binom{6}{4}=15=number\ of\ ways\ to\ choose\ 4\ red\ blocks

    \color{blue}\binom{7}{4}=35=number\ of\ ways\ to\ choose\ 4\ blue

    \color{red}\binom{6}{3}\color{blue}\binom{7}{1}=14  0=\color{red}3\ red\color{blue}\ with\ 1\ blue

    \color{red}\binom{6}{2}\color{blue}\binom{7}{2}=32  5=\color{red}2\ reds\color{blue}\ with\ 2\ blues

    \color{red}\binom{6}{1}\color{blue}\binom{7}{3}=21  0=\color{red}1\ red\color{blue}\ with\ 3\ blues

    total=715

    (ii)

    Why do you think your answer is wrong??

    It was ok to be suspicious, however \binom{13}{4}=715

    You get the same answer by summing the individual cases.


    (iii)

    the purpose of the question is for you to discover that there is a fast way to calculate (iii)
    At least 1 is blue is equivalent to "all 4 are not red".

    Since all probabilities sum to 1,
    the answer is

    1-\color{red}P(all\ 4\ are\ red)\color{black}=1-\frac{\color{red}\binom{6}{4}\color{black}}{\binom  {13}{4}}=1-\frac{\color{red}15}{75}
    thanks! good explanation
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