Consider rolling two fair 6-sided dice. We record the results as (n_1, n_2) with
ID: 3172504 • Letter: C
Question
Consider rolling two fair 6-sided dice. We record the results as (n_1, n_2) with n_1 being the value showing on die 1, and n_2 the value showing on die 2. What is the sample space ohm, and how large is this space (i.e., |ohm|)? What would be a reasonable probability measure for events E ohm, i.e., how would you compute the P(E) Let A = {die 1 shows an odd value}. Find P(A) Let B {die 2 shows an odd value}. Find P(B). Let C = the sum of the values on the two dice is odd}. Find P(C). Find P(A intersection B), P(A intersection C), and P(B intersection C). What does this say about the events? Find P(A intersection B intersection C). What does this say about the events? Are they mutually independent?Explanation / Answer
Solution :-
(a) The sample space S for this experiment is S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
II = 36
(b) P(E) = 1/36 + 1/36 + .... = 1
(c) A = {die 1 shows an odd value} = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) }
P(A) = 18 / 36 = 0.5
(d) B = {die 2 shows an odd value} = { (1, 1), (1, 3), (1, 5), (2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5), (5, 1), (5, 3), (5, 5), (6, 1), (6, 3), (6, 5) }
P(B) = 18 / 36 = 0.5
(e) C = {sum of the values on the two dice is odd} = { (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5) }
P(C) = 18 / 36 = 0.5
(f) P(A and B) = 9 / 36 = 1/4 (That is an event in A and B both, like (1, 1) or (3, 1), etc.)
P(A and C) = 9 / 36 = 1/4
P(B and C) = 9 / 36 = 1/4
These events are not mutually exclusive.
(g) P(A and B and C) = 0. This shows that no event is common between A, B and C. They are mutually independent.
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