instructions. show all your work to get full credit. circle your final answers. the exam is worth 40 points…

instructions. show all your work to get full credit. circle your final answers. the exam is worth 40 points that is 20% of your final grade. 1) (4 points) suppose given the sample space s = {e1,e2,e3}. indicate the cases where we have an acceptable assignment of probabilities to the simple events. explain the reasons if an assignment is unacceptable. (a) p(e1) = 1/4; p(e2) = 1/3; p(e3) = 1/4. (b) p(e1) = 1/3; p(e2) = -1/3; p(e3) = 1. (c) p(e1) = 1/5; p(e2) = 3/5; p(e3) = 1/5. (d) p(e1) = 1/2; p(e2) = 0; p(e3) = 1/2.

instructions. show all your work to get full credit. circle your final answers. the exam is worth 40 points that is 20% of your final grade. 1) (4 points) suppose given the sample space s = {e1,e2,e3}. indicate the cases where we have an acceptable assignment of probabilities to the simple events. explain the reasons if an assignment is unacceptable. (a) p(e1) = 1/4; p(e2) = 1/3; p(e3) = 1/4. (b) p(e1) = 1/3; p(e2) = -1/3; p(e3) = 1. (c) p(e1) = 1/5; p(e2) = 3/5; p(e3) = 1/5. (d) p(e1) = 1/2; p(e2) = 0; p(e3) = 1/2.

Answer

Explanation:

Step1: Recall probability axioms

The sum of probabilities of all simple - events in a sample space must be equal to 1, and the probability of any event (P(E)) must satisfy (0\leq P(E)\leq1).

Step2: Analyze part (a)

For the sample space (S = {e_1,e_2,e_3}), (P(e_1)=\frac{1}{4}), (P(e_2)=\frac{1}{3}), (P(e_3)=\frac{1}{4}). Calculate the sum (P(e_1)+P(e_2)+P(e_3)=\frac{1}{4}+\frac{1}{3}+\frac{1}{4}=\frac{3 + 4+3}{12}=\frac{10}{12}=\frac{5}{6}\neq1). So, this is not a valid assignment of probabilities.

Step3: Analyze part (b)

(P(e_1)=\frac{1}{3}), (P(e_2)=-\frac{1}{3}), (P(e_3) = 1). Since (P(e_2)=-\frac{1}{3}<0), this is not a valid assignment of probabilities because probabilities must be non - negative.

Step4: Analyze part (c)

(P(e_1)=\frac{1}{5}), (P(e_2)=\frac{3}{5}), (P(e_3)=\frac{1}{5}). Calculate the sum (P(e_1)+P(e_2)+P(e_3)=\frac{1 + 3+1}{5}=1), and (0\leq P(e_i)\leq1) for (i = 1,2,3). So, this is a valid assignment of probabilities.

Step5: Analyze part (d)

(P(e_1)=\frac{1}{2}), (P(e_2)=0), (P(e_3)=\frac{1}{2}). Calculate the sum (P(e_1)+P(e_2)+P(e_3)=\frac{1}{2}+0+\frac{1}{2}=1), and (0\leq P(e_i)\leq1) for (i = 1,2,3). So, this is a valid assignment of probabilities.

Answer:

(a) Not a valid probability assignment because (P(e_1)+P(e_2)+P(e_3)\neq1). (b) Not a valid probability assignment because (P(e_2)<0). (c) Valid probability assignment as (P(e_1)+P(e_2)+P(e_3)=1) and (0\leq P(e_i)\leq1) for (i = 1,2,3). (d) Valid probability assignment as (P(e_1)+P(e_2)+P(e_3)=1) and (0\leq P(e_i)\leq1) for (i = 1,2,3).