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Math Help - probability and geometry

  1. #1
    Junior Member
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    probability and geometry

    help, help, help........................
    can you guys help me on these two problems thanks especially the first

    1. Three semicircles of radius 1 are constructed on diameter of a semicircle of radius 2. The centers of the small semicircles divide into four line segments of equal length, as shown. What is the exact area of the shaded region that lies within the large semicircle but outside the smaller semicircles?

    picture udnerneath

    2. A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?
    a. (.3)
    b. (.4)
    c. (.5)
    d. (.6)
    e. (.7)

    thanks
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, anime_mania!

    2. A box contains exactly five chips, three red and two white.
    Chips are randomly removed one at a time without replacement
    until all the red chips are drawn or all the white chips are drawn.
    What is the probability that the last chip drawn is white?

    (a)\;0.3\qquad (b)\;0.4\qquad(c)\;0.5\qquad(d)\;0.6\qquad(e)\;0.7
    There are 6 outcomes in which the last chip is white.
    The cases are listed below with their corresponding probabilities.

    . . \begin{array}{cccc} \text{RRWW:} & \frac{3}{5}\cdot\frac{2}{4}\cdot\frac{2}{3}\cdot\f  rac{1}{2} & =& \frac{1}{10} \\ \\<br /> <br />
\text{RWRW:} & \frac{3}{5}\cdot\frac{2}{4}\cdot\frac{2}{3}\cdot\f  rac{1}{2} & =& \frac{1}{10} \\ \\<br /> <br />
\text{RWW:} & \frac{3}{5}\cdot\frac{2}{4}\cdot\frac{1}{3} &=& \frac{1}{10} \end{array}

    . . \begin{array}{cccc}\text{WRRW:} & \frac{2}{5}\cdot\frac{3}{4}\cdot\frac{2}{3}\cdot\f  rac{1}{2} &=& \frac{1}{10} \\ \\<br />
\text{WRW:} & \frac{2}{5}\cdot\frac{3}{4}\cdot\frac{1}{3} &=& \frac{1}{10} \\ \\<br /> <br />
\text{WW:} & \frac{2}{5}\cdot\frac{1}{4} &=& \frac{1}{10}<br />
\end{array}


    Hence: . P(\text{last is white}) \;=\;\frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} \;=\;\frac{6}{10}

    Therefore: . {\bf(d)\;\;0.6}

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