Answered

At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Join our platform to connect with experts ready to provide precise answers to your questions in different areas. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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
dgsistemasp0pffb

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!

We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.