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Consider a sample space of three outcomes A,B,and C. Which of the following represent legitimate probability models? 
 A. P(A)=0.1, P(B)=0.1, P(C)=1 
 B. P(A)=0.2, P(B)=0.1, P(C)=0.7 
 C. P(A)=-0.4, P(B)=0.9, P(C)=0.5 
 D. P(A)=0.5, P(B)=0,P(C)=0.4 
 E. P(A)=0.3, P(B)=0.3, P(C)=0.4

Sagot :

answer B & C are correct because P(A) + P(B) + P(C) add up to 1
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Answer:

B. P(A)=0.2, P(B)=0.1, P(C)=0.7

E. P(A)=0.3, P(B)=0.3, P(C)=0.4

Step-by-step explanation:

A fundamental property of probability is that the sum of the probabilities of the events is equal to 1.

Another property is that the probability of any event must be greater than or equal to zero, it can never be a negative value.

Case A. P(A)=0.1, P(B)=0.1, P(C)=1

P(A) + P(B) + P(C) = 0.1 + 0.1 + 1 = 1.2 ≠ 1 (It is not a legitimate probability model)

Case B. P(A)=0.2, P(B)=0.1, P(C)=0.7

P(A) + P(B) + P(C) = 0.2 + 0.1 + 0.7 = 1 (It is a legitimate probability model)

Case C. P(A)=-0.4, P(B)=0.9, P(C)=0.5

P(B) = -0.4 is a negative value (It is not a legitimate probability model)

Case D. P(A)=0.5, P(B)=0,P(C)=0.4

P(A) + P(B) + P(C) = 0.5 + 0 + 0.4 = 0.9 ≠ 1 (It is not a legitimate probability model)

Case E. P(A)=0.3, P(B)=0.3, P(C)=0.4

P(A) + P(B) + P(C) = 0.3 + 0.3 + 0.4 = 1 (It is a legitimate probability model)

Hope this helps!

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