Suppose that one die is rolled and you observe the number of dots facing up. The

Question

Suppose that one die is rolled and you observe the number of dots facing up. The sample space for this experiment is = {1, 2, 3, 4, 5, 6}. The following table provides 5 different potential probability assignments to the possible outcomes. Outcome Assignment #1 Assignment #2 Assignment #3 Assignment #4 Assignment #5 1 0 1 8 1 6 1 2 0.3 2 1 4 1 4 1 6 1 2 0.13 3 1 8 7 16 1 6 1 4 0.08 4 1 16 1 16 1 6 1 4 0.1 5 1 8 1 8 1 6 0.5 0.22 6 7 16 1 4 1 6 0.5 0.17 a) For each of the assignments #1-#5, state whether it is a legitimate probability assignment or not. If not, explain why it is not legitimate. Answer parts b), c) and d) using only the legitimate probability assignments from the table. (i.e. only use the assignments you said were legitimate in part a). b) Let A = die comes up even; B = die comes up at least 2; C = die comes up at most 5; D = die comes up 1. Determine the probability of each of these four events using each of the legitimate probability assignments. c) Determine the probability that you observe a die roll which is a multiple of 3. d) Determine the probability that you observe a prime number on a die roll (note: one is not a prime number) e) If the die is balanced (i.e. fair), which probability assignment would be appropriate? Why? f) Using probability assignment #5 and the events defined in part b) above, find P(A B), P(B C), P(B C), P(A C D), P(A C) C , P(D|B), and P(A (B C))

Explanation / Answer

P(A B) = P(die comes even and die comes atleast 2

= P( die shows 2,4 or 6)

= 0.5

P(B C) = P(atleast 2,and atmost 5)

= P(2 or 3 or 4 or 5) = 2/3

P(B C) = P(2 or 3 or 4 or 5 or 6 or 1)

= 1

P(A C D) = P( all A, C, D together)

= P( 2 or 4 and 1)

=0

P(A C) C = P(C) = 5/6

P(D|B) = P(BD)/P(B) = 0 (impossible event)

P(A (B C)) = P(A) + P(BC)-P(ABC)

= 1/2+ 2/3 - P(2,4)

= 1/2+2/3-1/3

= 5/6

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