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If an experiment has the three possible and mutually exclusive outcomes $A, B$, and $C$, check in each case whether the assignment of probabilities is permissible:

(a) $P(A)=\frac{1}{3}, P(B)=\frac{1}{3}$, and $P(C)=\frac{1}{3}$
(b) $P(A)=0.64, P(B)=0.38$, and $P(C)=-0.02$
(c) $P(A)=0.35, P(B)=0.52$, and $P(C)=0.26$
(d) $P(A)=0.57, P(B)=0.24$, and $P(C)=0.19$
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## 1 Answer

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(a) The assignment of probabilities is permissible because the values are all on the interval from 0 to 1 , and their sum is $\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$.

(b) The assignment is not permissible because $P(C)$ is negative.

(c) The assignment is not permissible because $0.35+0.52+0.26=1.13$, which exceeds $1 .$

(d) The assignment is permissible because the values are all on the interval from 0 to 1 and their sum is $0.57+0.24+0.19=1$.
by Diamond (40,719 points)

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