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Math Help - Probabilities - more questions

  1. #1
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    Probabilities - more questions

    3. A bag contains 10 balls, 4 are red and 6 are yellow. Three balls are drawn without replacement. What is the probability of drawing at least two red balls?


    4. A town has two fire engines operating independently. The probability that a specific fire engine is available when need is 0.95. What is the probability that neither is available when needed? What is the probability that a fire engine is available when needed?



    OK could someone help out a little bit my brain is failing to work, I have done some working out but none of it seems to make sense and it seems easy. Don't know what's going on up there.
    Last edited by mr fantastic; December 15th 2009 at 06:40 PM. Reason: Moved from another thread.
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    What is the working out you have done?
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  3. #3
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    Hello, djmccabie!

    3. A bag contains 10 balls, 4 are red and 6 are yellow.
    Three balls are drawn without replacement.
    What is the probability of drawing at least two red balls?

    We draw 3 balls from 10 balls.
    There are: . {10\choose3} = 120 possible outcomes.


    We want 3 Red balls, or 2 Red and 1 Yellow.

    . . 3R: . {4\choose3} = 4 ways.
    . . 2R, 1Y: . {4\choose2}{6\choose1} = 6\cdot6 = 36 ways.

    Hence, there are: . 4 + 36 \:=\:40 ways to get at least 2 red balls.


    Therefore: . P(\text{at least 2 R}) \;=\;\frac{40}{120} \;=\;\frac{1}{3}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    You could have solved it taking baby steps . . .


    There are four ways to get at least 2 red balls: .RRR, RRY, RYR, YRR.

    . . \begin{array}{ccccc}P(RRR) &=& \frac{4}{10}\cdot\frac{3}{9}\cdot\frac{2}{8} &=& \frac{1}{30} \\ \\[-3mm]<br />
P(RRY) &=&\frac{4}{10}\cdot\frac{3}{9}\cdot\frac{6}{8} &=& \frac{1}{10} \\ \\[-3mm]<br />
P(RY\!R) &=& \frac{4}{10}\cdot\frac{6}{9}\cdot\frac{3}{8} &=& \frac{1}{10} \\ \\[-3mm]<br />
P(Y\!RR) &=& \frac{6}{10}\cdot\frac{4}{9}\cdot\frac{3}{8} &=& \frac{1}{10}\end{array}

    Therefore: . P(\text{at least 2 R}) \;=\;\frac{1}{30} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} \;=\;\frac{10}{30} \;=\;\frac{1}{3}

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