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# How do you determine the required value of the missing probability to make the following distribution a discrete probability distribution?

## X P(x) 0 0.30 1 0.15 2 ? 3 0.20 4 0.15 5 0.05

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#### Explanation

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#### Explanation:

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9
Feb 6, 2017

$P \left(2\right) = 0.15$.

#### Explanation:

In any Prob. Dist., we must have, (1) sumP(x)=1, &, (2)" each "P(x)>=0.

Now, $\sum P \left(x\right) = 1 \Rightarrow P \left(0\right) + P \left(1\right) + P \left(2\right) + \ldots + P \left(5\right) = 1$

$\Rightarrow 0.30 + 0.15 + P \left(2\right) + 0.20 + 0.15 + 0.05 = 1$

$\Rightarrow P \left(2\right) = 0.15$.

We readily have, each$P \left(x\right) \ge 0.$

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