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Math Help - Impossible... I cant get this.. If any one else can There a genius..

  1. #1
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    Impossible... I cant get this.. If any one else can There a genius..

    7. A drawer contains four red socks and five blue socks.
    (a) Three socks are drawn one at a time and then put back before the next
    selection. Determine the probability that
    (i) exactly two red socks are selected
    (ii) at least two red socks are selected
    (b) Repeat part (a) without replacement

    answeres... ia) 80over243 ib) 304over729
    iia) 5over14 iib) 17over42

    (over is a fraction)
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by wikji View Post
    7. A drawer contains four red socks and five blue socks.
    (a) Three socks are drawn one at a time and then put back before the next
    selection. Determine the probability that
    (i) exactly two red socks are selected
    (ii) at least two red socks are selected
    (b) Repeat part (a) without replacement

    answeres... ia) 80over243 ib) 304over729
    iia) 5over14 iib) 17over42

    (over is a fraction)
    to pick one red sock

    \frac{\binom{4}{1}}{\binom{9}{1}}


    to pick one blue sock

    \frac{\binom{5}{1}}{\binom{9}{1}}

    Since we are replacing the socks the events are independant so we multiply the probibilites together

    \frac{\binom{4}{1}}{\binom{9}{1}}\cdot \frac{\binom{4}{1}}{\binom{9}{1}} \cdot \frac{\binom{5}{1}}{\binom{9}{1}}=\frac{80}{729}

    since there are three different ways we can draw 2 red and 1 blue {r,r,b} or {r,b,r} or {b,r,r} we multiply our prob by 3

    \frac{80}{729} \cdot 3 =\frac{80}{243}

    for two use the same process to find the prob of 3 red socks. then add the answer from part 1 to this answer.

    Without replacement

    \frac{\binom{4}{2}  \binom{5}{1}}{\binom{9}{3}}=\frac{5}{14}

    use the same as above for the 2nd part
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