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Math Help - Jar with 5R&2G balls

  1. #1
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    Unhappy Jar with 5R&2G balls

    Okay, last one!

    You have a jar with 5Red balls & 2Green balls--a ball's chosen randomly. It's returned to the jar along w/6 more of the same color. Then, a ball is drawn again--calculate the probability that the second ball is Red, as a fraction in lowest terms, showing all intermediate steps.

    Also, find the prob. that both balls drawn are same color--don't need to simplify this answer.

    THANK YOU SO MUCH!!
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  2. #2
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    Hello,
    hope this is correct..

    Pr(\mbox{Second is red})=P(R|G)+P(R|R)=\frac{2}{7}\cdot \frac{5}{13} + \frac{5}{7}\cdot \frac{11}{13}=\frac{5}{7}

    Pr(\mbox{Both the same colour})=P(R|R)+P(G|G)=\frac{5}{7}\cdot \frac{11}{13}+\frac{2}{7}\cdot \frac{8}{13}=\frac{71}{91}
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  3. #3
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    Hello, bjanela!

    You're right, Robb!

    I'll baby-step through it . . .


    You have a jar with 5 Red balls and 2 Green balls.
    A ball is chosen randomly.
    It's returned to the jar along w/6 more of the same color.
    Then, a ball is drawn again.

    (a) Find the probability that the second ball is Red.
    There are two cases to consider:
    . . [1] The first ball is Red.
    . . [2] The first ball is Green.


    Case 1: the first ball is red.
    . . P(\text{1st Red}) \:=\:\frac{5}{7}

    Then the jar has: .11 Red, 2 Green
    . . P(\text{2nd Red}) \:=\:\frac{11}{13}

    Hence: .= P(\text{1st Red}\:\wedge\:\text{2nd Red}) \:=\:\frac{5}{7}\cdot\frac{11}{13} \:=\:\frac{55}{91}


    Case 2: the first ball is green.
    . . P(\text{1st Green}) \:=\:\frac{2}{7}

    Then the jar has: .5 Red, 8 Green
    . . P(\text{2nd Red}) \:=\:\frac{5}{13}

    Hence: . P(\text{1st Green}\:\wedge\:\text{2nd Red}) \:=\:\frac{2}{7}\cdot\frac{5}{13} \:=\:\frac{10}{91}


    Therefore: . P(\text{2nd Red}) \;=\;\frac{55}{91} + \frac{10}{91} \;=\;\frac{65}{91} \;=\;\boxed{\frac{5}{7}}



    (b) Find the probabiity that both balls drawn are same color.

    We already know: . P(\text{1st Red}\:\wedge\:\text{2nd Red}) \;=\;\frac{55}{91}

    We have: . P(\text{1st Green}) \:=\:\frac{2}{7}

    . . The jar has: .5 Red, 8 Green

    . . Then: . P(\text{2nd Green}) \:=\:\frac{8}{13}

    . . Hence: . P(\text{1st Green}\:\wedge\:\text{2nd Green}) \:=\:\frac{2}{7}\cdot\frac{8}{13} \:=\:\frac{16}{91}


    Therefore: . P(\text{same color}) \;=\;\frac{55}{91} + \frac{16}{91} \;=\;\boxed{\frac{71}{91}}

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