Problem 2. (30 points) a) (5 points) In rolling 3 fair dice, what is the probabi

Question

Problem 2. (30 points) a) (5 points) In rolling 3 fair dice, what is the probability of obtaining a sum not greater than 7? b) (5 points) In rolling 2 fair dice, what is the probability of a sum greater than 3 but not exceeding 6? c) (5 points) Given that the first roll was an odd number what is the probability that sum exceeds 6? The notation for this is P(A B) Probability(sum exceeds 6 given that the first roll was 1, 3 or 5) d) (5 points) The definition for independent events: two events, A and 8, are independent if P(A I 8) P(A). Given events A sum exceeds 6, and B first roll was an odd num ber, are they independent? In other words, (one of the student's words): Given that the first die was an odd number (1, 3, 5), what is the probability that the sum exceeds 6 when rolling the second die. e) (5 points) What if events A both rolls are the same, and 8-first roll was an odd number? (5 points) Are the results from d and e surprising? Interpret.

Explanation / Answer

a) in each dice numbers from 1 to 6 can be possible...for sum not greater than 7 , means sum<=7
Minimum sum possible for 3 dices is 3(1,1,1 on each dice)
Possible cases(1,1,1)(1,2,1)(1,1,2),(2,1,1)(3,1,1)(1,3,1)(1,1,3)(2,2,1)(2,1,2)(1,2,2)(4,1,1)(1,4,1)(1,1,4)(3,2,1)(3,1,2)(1,2,3)(2,1,3)(2,3,1)(1,3,2)(2,2,2)(5,1,1)(1,5,1)(1,1,5)(4,2,1)(4,1,2)(1,4,2)(2,4,1)(1,2,4)(2,1,4)(3,3,1)(3,1,3)(1,3,3)(3,2,2)(2,3,2)(2,2,3) =35 cases
Probability =35/(6^3) =35/216

b)for sum greater than 3 but not exceeding 6...so possible sum are 4 and 5
For sum 5 =(1,4)(4,1)(2,3)(3,2)
For sum 4 =(1,3)(3,1)(2,2)
So total 7 cases
Probability =7/36
36 because total cases will be 6*6 for 2 fair dices.

C) P(A|B) =P(A?B)/P(B)
For P(A?B) cases are (1,6)(3,4)(3,5)(3,6)(5,2)(5,3)(5,4)(5,5)(5,6)
So P(A?B)=9/36= 1/4
So for P(B), all 6 cases for each 1,3,5 so total 18
P(B)=18/36=1/2
So P(A|B) =1/2

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