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A random variable $Y$ can only take values in $\{−10, 0, 10\}$. The expected value of $Y$ is 0 and its variance is 80. Find the probability distribution of $Y$.

Knowing the expected value and variance enables me to decide either this is binomial or geometric but since they want probability distribution, I need some understanding on how to solve this question.

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Let the probabilities of $Y=-10,0,10$ be $a,b,c$ respectively. Then obviously we have $a+b+c=1$, but since the expected value is zero we have $-10a+10c=0$ or $a=c$. The variance of 80 leads to the third equation $$a(-10)^2+b(0^2)+c(10)^2=80$$ $$100(a+c)=80$$ $$a=c=\frac{80}{2\cdot100}=\frac25$$ Then $b=1-\frac25-\frac25=\frac15$.

$Y$ thus takes 0 with probability $\frac15$, $+10$ with probability $\frac25$ and $-10$ with probability $\frac25$.

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