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Math Help - Die thrown 3 times. If odd prime on first throw .....

  1. #1
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    Die thrown 3 times. If odd prime on first throw .....

    a man throws a die three times .if in first throw an odd prime no occurs he adds the numbers appearing in the next two draws otherwise he multiplies them.find the probability that the number thus obtained is even number

    i am trying to use binomial distribution
    is it right
    Last edited by mr fantastic; August 3rd 2011 at 07:13 PM. Reason: Re-titled.
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    Re: probability

    Hello, prasum!

    A man throws a die three times.
    If in first throw, an odd prime number occurs
    . . he adds the numbers appearing in the next two throws.
    Otherwise, he multiplies them.

    Find the probability that the number thus obtained is even number.

    The first throw is a prime: {3, 5}. . P(\text{prime}) = \tfrac{2}{6} = \tfrac{1}{3}

    The sum of the next two throws is even.
    . . . . P(\text{even and even}) \:=\:\tfrac{3}{6}\cdot\tfrac{3}{6} \:=\:\tfrac{1}{4}
    . . . . P(\text{odd and odd}) \:=\:\tfrac{3}{6}\cdot\tfrac{3}{6}\:=\:\tfrac{1}{4  }
    So: P(\text{sum is even}) \:=\:\tfrac{1}{4} + \tfrac{1}{4} \:=\:\tfrac{1}{2}

    Hence: P(\text{prime }\wedge\text{ even sum}) \:=\:\tfrac{1}{3}\cdot\tfrac{1}{2} \:=\:\tfrac{1}{6}


    The first throw is not a prime: {1, 2, 4, 6}. . P(\sim\text{prime}) = \tfrac{4}{6} = \tfrac{2}{3}

    The product of the next two throws is odd.
    . . P(\text{odd and odd}) \:=\:\tfrac{3}{6}\cdot\tfrac{3}{6} \:=\:\tfrac{1}{4}
    So: P(\text{even product}) \:=\:1 - \tfrac{1}{4} \:=\:\tfrac{3}{4}

    Hence: P(\sim\text{prime }\wedge\text{ even product}) \:=\:\tfrac{2}{3}\cdot\tfrac{3}{4}\:=\:\tfrac{1}{2  }


    Therefore: P(\text{even}) \:=\:\tfrac{1}{6} + \tfrac{1}{2} \:=\:\tfrac{2}{3}

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    Re: probability

    Quote Originally Posted by prasum View Post
    a man throws a die three times .if in first throw an odd prime no occurs he adds the numbers appearing in the next two draws otherwise he multiplies them.find the probability that the number thus obtained is even number

    i am trying to use binomial distribution
    is it right
    I suggest you construct two probabilty grids to answer this question - one for if the first thrown number is an odd prime and one for if it's not. Note that Pr(Odd prime) = 1/3.
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