8. Suppose a pair of six sided dice are rolled, and the numbers on the upper mos

Question

8. Suppose a pair of six sided dice are rolled, and the numbers on the upper most faces are recorded.
(a) Give the sample space S for this experiment.
(b) What is the probability that the total of the numbers on the uppermost faces total 7?
(c) What is the probability that the total of the numbers on the uppermost faces total 2?
(d) What is the probability that the total of the numbers on the uppermost faces total 1?
(e) What is the probability that the total of the numbers on the uppermost faces is 8 or that at least one
of the numbers is 6?

Explanation / Answer

(a) Sample space = the set of all possible outcomes. For rolling two dice, it would be x,y where x and y are each {1,2,3,4,5,6} Thus, if 32 represents rolling a 3 on the first die and a 2 on the second die, the sample space would be S= {11,12,13,14,15,16, 21,22,23,24,25,26, 31,32,33,34,35,36, 41,42,43,44,45,46, 51,52,53,54,55, 61,62,63,64,65,66}. (b) The probability that the total of the numbers on the uppermost faces is 7 would be the number of events in the sample space where the two numbers add up to 7 divided by the number of events in the sample space. The events where the numbers add up to 7 are {16,25,34,43,52,61}. There are six of these, and 36 events in the sample space, so the probability is 6/36 = 1/6. (c) The probability that the total of the numbers on the uppermost faces is 2 would be the number of events in the sample space where the two numbers add up to 2 divided by the number of events in the sample space. There is only one event in the sample space in which the numbers add up to 2. It would be {11}, so the probability is 1/36. (d) The probability that the total of the numbers on the uppermost faces is 1 would be the number of events in the sample space where the two numbers add up to 1 divided by the number of events in the sample space. There are no events in the sample space in which the numbers add up to 1, so the probability is 0/36 = 0. (e) To find the probability that the total of the numbers on the uppermost faces is 8 or that at least on of the numbers is 6, we count the events that meet either of these qualifications. They would be {26,35,44,53,62} U {16,26,36,46,56,61,62,63,64,65,66} = {16,26,35,36,44,46,53,56, 61,62,63,64,65,66} There are 14 elements in this subset, so the probability is 14/36 = 7/18.

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