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

Question

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

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Answer 1 · 2017-02-06T17:46:26

# P(2)=0.15#.

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.#