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An Urn contains 7 red and 11 white balls. Draw one ball at random from the urn. Let X=1 if a red ball is drawn, and let X=-1 if a white ball is drawn. Give the pmf, mean, and Variance of X.

I know how to give the pmf, mean, and variance if X=0 or 1, but the -1 throws me off. How would i go about doing this? Thank you for any help!

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Just use the formulas as you would in any other given scenario...

PMF:

  • $P(X=+1)=\dfrac{7}{18}$
  • $P(X=-1)=\dfrac{11}{18}$

Mean: $1\cdot\dfrac{7}{18}-1\cdot\dfrac{11}{18}=-\dfrac{2}{9}$

Variance: $(1+\dfrac{2}{9})^2+(-1+\dfrac{2}{9})^2=\dfrac{170}{81}$

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Hint: If $Y=2X-1$ then:

  • $\Pr(Y=2x-1)=\Pr(2X-1=2x-1)=\Pr(X=x)$
  • $E[Y]=E[2X-1]= 2E[X]-1$
  • $Var[Y]=Var[2X-1]=Var[2X]+Var[1]=2^2Var[X]+0=4Var[X]$

since $X$ and $1$ are independent and $1$ is a constant.

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PMF: \begin{equation} f(x)=\begin{cases} 7/18, & \text{if $x=1$}.\\ 11/18, & \text{if $x=-1$}. \end{cases} \end{equation}

Mean: $$E(X) = x_1P(x_1) + x_2P(x_2) = 1 * 7/18 - 1 * 11/18 = -2/9$$

Variance:

$$\sigma^2 = (x_1 - E(X))^2 * P(x_1) + (x_2 - E(X))^2 * P(x_2) = (11/9)^2 * 7/18 + (7/9)^2 * 11/18 = 77/81$$

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