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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

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

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9
Feb 6, 2017

Answer:

# P(2)=0.15#.

Explanation:

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

Now, #sumP(x)=1 rArr P(0)+P(1)+P(2)+...+P(5)=1#

#rArr 0.30+0.15+P(2)+0.20+0.15+0.05=1#

#rArr P(2)=0.15#.

We readily have, each#P(x)>=0.#

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