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Math Help - Constructing a p.m.f.

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Constructing a p.m.f.

    I'm still having a bit of trouble trying to understand this stuff.

    Let a chip be taken at random from a bowl that contains six white chips, three red chips, and one blue chip. Let the random variable X=1 if the outcome is a white chip; let X=5 if the outcome is a red chip; and left X=10 if the outcome is a blue chip.

    (a) Find the p.m.f. of X
    [snip]
    Source: Probability and Statistical Inferences, 7E, by Hoggs and Tanis

    I understand that P(X=1)=\frac{6}{10}, P(X=5)=\frac{3}{10}, and P(X=10)=\frac{1}{10}.

    My issue here is determining a proper value for the numerator of my f(x)=P(X=x). My stab at this would be to say that the p.m.f. has the form of f(x)=\frac{u}{10}, where u is the part I can't figure out.

    I see a pattern though:

    X=1:~~~~~6

    X=5:~~~~~3

    X=10:~~~~\!1

    The difference between the first two terms is 3, and the last two terms is 2. Other than that, I'm at a standstill.

    I'd appreciate any input!

    --Chris



    w00t!!! my 9th post!!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Chris L T521 View Post
    I'm still having a bit of trouble trying to understand this stuff.

    Source: Probability and Statistical Inferences, 7E, by Hoggs and Tanis

    I understand that P(X=1)=\frac{6}{10}, P(X=5)=\frac{3}{10}, and P(X=10)=\frac{1}{10}.

    My issue here is determining a proper value for the numerator of my f(x)=P(X=x). My stab at this would be to say that the p.m.f. has the form of f(x)=\frac{u}{10}, where u is the part I can't figure out.

    I see a pattern though:

    X=1:~~~~~6

    X=5:~~~~~3

    X=10:~~~~\!1

    The difference between the first two terms is 3, and the last two terms is 2. Other than that, I'm at a standstill.

    I'd appreciate any input!

    --Chris



    w00t!!! my 9th post!!
     <br />
f(x) = \begin{cases} <br />
\frac{6}{10}, &x\in \{1\},\\<br />
\frac{3}{10}, &x\in \{5\},\\<br />
\frac{1}{10}, &x\in \{10\},\\<br />
0, &x\in \mathbb{R}\backslash \{1,\ 5,\ 10\}.\end{cases}<br />

    RonL
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