# What does this distribution mean?

I'm kind of embarrassed that I don't fully get this, but I assume it is because the example is using letters for a sample space instead of numbers.

Sample space = {a,b,c,d,e,f} with equal probability

outcome |  a  |  b  |  c  |  d  |  e  |  f  |
x       |  0  |  0  | 1.5 | 1.5 |  2  |  3  |


Is this saying that P(X=0) = a + b? Or P(X=0) = 1/6 + 1/6?

Could someone please explain this is simple terms (like a dice throw example, perhaps)?

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You did not say what the question was. Presumably the random variable $X$ is defined to be $0$ at $a$ and $b$, $1.5$ at $c$, and so on. And we are asked a question about $X$, such as the mean of $X$. But without the question it is hard to discuss details. – André Nicolas Oct 9 '12 at 23:08
I've asked a concept question, so there isn't really a question. But I guess finding P(X=1.5) would help me understand. – n0pe Oct 9 '12 at 23:10
$1/6+1/6$.${}{}{}{}$ – André Nicolas Oct 9 '12 at 23:12
There are six equally likely outcomes. Think of them as the outcomes of a roll of a fair die instead of a, b, c, d, e, f if you wish. Suppose you win an amount $X$ whose value depends on the outcome of the die roll. You win nothing if you roll a 1 or a 2, 1.5 if you roll a 3 or a 4, 2 if you roll a 5, and 3 if you roll a 6. So $X$ takes on values 0, 1.5, 2, and 3 with probabilities $\frac{1}{6}+\frac{1}{6} = \frac{1}{3}$, $\frac{1}{6}+\frac{1}{6} = \frac{1}{3}$, $\frac{1}{6}$ and $\frac{1}{6}$ respectively, – Dilip Sarwate Oct 9 '12 at 23:13
@DilipSarwate, this is exactly what I needed, thank you. If you're willing to post it as an answer, I'll mark it as accepted. – n0pe Oct 9 '12 at 23:15

## 1 Answer

There are six equally likely outcomes a, b, c, d, e, f. Think of them as the outcomes of a roll of a fair die whose faces are marked a, b, c, d, e, f instead of the usual $1, 2, 3, 4, 5, 6$. Suppose you win an amount $X$ whose value depends on the outcome of the die roll. The table shows that you win $0$ if you roll an a or a b, $1.5$ if you roll a c or a d, $2$ if you roll an e, and $3$ you roll an f. So X takes on values $0, 1.5, 2$, and $3$ with probabilities $\frac{1}{6}+\frac{1}{6} = \frac{1}{3}$, $\frac{1}{6}+\frac{1}{6} = \frac{1}{3}$, $\frac{1}{6}$ and $\frac{1}{6}$ respectively,

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