Constructing a p.m.f.

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• Sep 23rd 2008, 09:39 PM
Chris L T521
Constructing a p.m.f.
I'm still having a bit of trouble trying to understand this stuff.

Quote:

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 9(Sun)(Sun)th post!! :D
• Sep 24th 2008, 12:00 AM
CaptainBlack
Quote:

Originally Posted by Chris L T521
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 9(Sun)(Sun)th post!! :D

$
f(x) = \begin{cases}
\frac{6}{10}, &x\in \{1\},\\
\frac{3}{10}, &x\in \{5\},\\
\frac{1}{10}, &x\in \{10\},\\
0, &x\in \mathbb{R}\backslash \{1,\ 5,\ 10\}.\end{cases}
$

RonL