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Math Help - Conditional probability?

  1. #1
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    Conditional probability?

    Two game titles numbered 1 through nine are selected at random from a box without replacement. If their sum is even what is the probability that both numbers are odd

    Well I know that even numbers are 2,4,6,8

    So my sums would be 2+2,3+1,4+2,3+3,6+2,5+1,7+1,4+4,7+1,4+4,1+1,5+3

    So I have 10 sums and 6 are with odd numbers.

    So I thought of doing 6/18 / 10/18 but apparently this is wrong.
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  2. #2
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    Re: Conditional probability?

    My second questions is two boys and two girls are lined up at random. What is the probability that the girls are separated if a girl is at an end?

    So these are my different sample spaces

    (B,B,G,G) (G,B,B,G) (B,G,G,B) (G,B,G,B) (B,G,B,G) (G,G,B,B)
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  3. #3
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    Re: Conditional probability?

    Quote Originally Posted by homeylova223 View Post
    Two game titles numbered 1 through nine are selected at random from a box without replacement. If their sum is even what is the probability that both numbers are odd
    The sum will be even if and only if the two are both even or both odd.
    So there are \binom{4}{2}+\binom{5}{2} ways to get an even sum.

    Now finish it.
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    Re: Conditional probability?

    The only thing I am confused on is that when I check the answer I get that the probability is 5/8? How can this be?
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    Re: Conditional probability?

    Quote Originally Posted by homeylova223 View Post
    The only thing I am confused on is that when I check the answer I get that the probability is 5/8? How can this be?
    What is \frac{\binom{5}{2}}{\binom{4}{2}+\binom{5}{2}}=~?
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  6. #6
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    Re: Conditional probability?

    Hello, homeylova223!

    Game titles numbered 1 through nine are in a box.
    Two are selected at random without replacement.
    If their sum is even, what is the probability that both numbers are odd?

    The very worst we can do is list the possible outcomes
    . . and count the desirable ones.


    . . \begin{array}{c}\text{Even sum} \\(1,3)\;(1,5)\;(1,7)\;(1,9) \\ (2,4)\;(2,6)\;(2,8) \\ (3,5)\;(3,7)\;(3,9) \\ (4,6)\;(4,8) \\ (5,7)\;(5,9) \\ (6,8) \\ (7,9) \end{array}\text{ 16 outcomes}


    . . \begin{array}{c}\text{Both odd} \\ (1,3)\;(1,5)\;(1,7)\;(1,9) \\ (3,5)\;(3.7)\;(3,9) \\ (5,7)\;(5,9) \\ (7,9) \end{array} \text{ 10 outcomes}


    \text{Therefore: }\:P(\text{both odd}\,|\,\text{even sum}) \;=\;\frac{10}{16} \;=\;\frac{5}{8}

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    Re: Conditional probability?

    For my second question would I do this


    6 ncr 3 As there are 3 girls who are seperated and 3 girls who are at the end in my sample space?
    --------
    6 ncr 3
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    Re: Conditional probability?

    Quote Originally Posted by homeylova223 View Post
    For my second question would I do this
    PLEASE state both questions in on thread OR create two different threads.
    I did not even see the second question.

    The given here mean we are only interested in rearrangements of BBGG end in a G~?

    So to answer this second question, how many ways are there to rearrange BBG so that the rearrangement does not end in G ?
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