In: Statistics and Probability
The following events are considered for a throw with two
dice:
A: The total of eyes is greater than 7.
B: Exactly one of the two numbers is a 5.
C: No 1 is rolled.
a) Calculate the probabilities
P (A), P (B), P (C), P (A ∩ B), P (A ∩ C), P (B ∩ C), P (A ∪ B), P
(A | B), P (A | C), P (C | A), P (B | C).
b) Are events A and B independent or disjoint?
(a) Total number of cases is 36. favorable cases are shown in bold.
(1,1) 2 | (2,1) 3 | (3,1) 4 | (4,1) 5 | (5,1) 6 | (6,1) 7 |
(1,2) 3 | (2,2) 4 | (3,2) 5 | (4,2) 6 | (5,2) 7 | (6,2) 8 |
(1,3) 4 | (2,3) 5 | (3,3) 6 | (4,3) 7 | (5,3) 8 | (6,3) 9 |
(1,4) 5 | (2,4) 6 | (3,4) 7 | (4,4) 8 | (5,4) 9 | (6,4) 10 |
(1,5) 6 | (2,5) 7 | (3,5) 8 | (4,5) 9 | (5,5) 10 | (6,5) 11 |
(1,6) 7 | (2,6) 8 | (3,6) 9 | (4,6) 10 | (5,6) 11 | (6,6) 12 |
P(A) = 15/36
(1,1) 2 | (2,1) 3 | (3,1) 4 | (4,1) 5 | (5,1) 6 | (6,1) 7 |
(1,2) 3 | (2,2) 4 | (3,2) 5 | (4,2) 6 | (5,2) 7 | (6,2) 8 |
(1,3) 4 | (2,3) 5 | (3,3) 6 | (4,3) 7 | (5,3) 8 | (6,3) 9 |
(1,4) 5 | (2,4) 6 | (3,4) 7 | (4,4) 8 | (5,4) 9 | (6,4) 10 |
(1,5) 6 | (2,5) 7 | (3,5) 8 | (4,5) 9 | (5,5) 10 | (6,5) 11 |
(1,6) 7 | (2,6) 8 | (3,6) 9 | (4,6) 10 | (5,6) 11 | (6,6) 12 |
P(B) = 10/36 = 5/18
(1,1) 2 | (2,1) 3 | (3,1) 4 | (4,1) 5 | (5,1) 6 | (6,1) 7 |
(1,2) 3 | (2,2) 4 | (3,2) 5 | (4,2) 6 | (5,2) 7 | (6,2) 8 |
(1,3) 4 | (2,3) 5 | (3,3) 6 | (4,3) 7 | (5,3) 8 | (6,3) 9 |
(1,4) 5 | (2,4) 6 | (3,4) 7 | (4,4) 8 | (5,4) 9 | (6,4) 10 |
(1,5) 6 | (2,5) 7 | (3,5) 8 | (4,5) 9 | (5,5) 10 | (6,5) 11 |
(1,6) 7 | (2,6) 8 | (3,6) 9 | (4,6) 10 | (5,6) 11 | (6,6) 12 |
P(C) = 11/36
(1,1) 2 | (2,1) 3 | (3,1) 4 | (4,1) 5 | (5,1) 6 | (6,1) 7 |
(1,2) 3 | (2,2) 4 | (3,2) 5 | (4,2) 6 | (5,2) 7 | (6,2) 8 |
(1,3) 4 | (2,3) 5 | (3,3) 6 | (4,3) 7 | (5,3) 8 | (6,3) 9 |
(1,4) 5 | (2,4) 6 | (3,4) 7 | (4,4) 8 | (5,4) 9 | (6,4) 10 |
(1,5) 6 | (2,5) 7 | (3,5) 8 | (4,5) 9 | (5,5) 10 | (6,5) 11 |
(1,6) 7 | (2,6) 8 | (3,6) 9 | (4,6) 10 | (5,6) 11 | (6,6) 12 |
P(A ∩ B) = 6/36 = 1/6
(1,1) 2 | (2,1) 3 | (3,1) 4 | (4,1) 5 | (5,1) 6 | (6,1) 7 |
(1,2) 3 | (2,2) 4 | (3,2) 5 | (4,2) 6 | (5,2) 7 | (6,2) 8 |
(1,3) 4 | (2,3) 5 | (3,3) 6 | (4,3) 7 | (5,3) 8 | (6,3) 9 |
(1,4) 5 | (2,4) 6 | (3,4) 7 | (4,4) 8 | (5,4) 9 | (6,4) 10 |
(1,5) 6 | (2,5) 7 | (3,5) 8 | (4,5) 9 | (5,5) 10 | (6,5) 11 |
(1,6) 7 | (2,6) 8 | (3,6) 9 | (4,6) 10 | (5,6) 11 | (6,6) 12 |
P (A ∩ C) = 0/36 = 0
(1,1) 2 | (2,1) 3 | (3,1) 4 | (4,1) 5 | (5,1) 6 | (6,1) 7 |
(1,2) 3 | (2,2) 4 | (3,2) 5 | (4,2) 6 | (5,2) 7 | (6,2) 8 |
(1,3) 4 | (2,3) 5 | (3,3) 6 | (4,3) 7 | (5,3) 8 | (6,3) 9 |
(1,4) 5 | (2,4) 6 | (3,4) 7 | (4,4) 8 | (5,4) 9 | (6,4) 10 |
(1,5) 6 | (2,5) 7 | (3,5) 8 | (4,5) 9 | (5,5) 10 | (6,5) 11 |
(1,6) 7 | (2,6) 8 | (3,6) 9 | (4,6) 10 | (5,6) 11 | (6,6) 12 |
P (B ∩ C) = 2/36 = 1/18
P (A ∪ B) = P(A) + P(B) - P(A ∩ B) = 24/36 = 2/3
P (A | B) = P(A ∩ B) / P(B) = 6/10 = 0.6
P (A | C) = P(A ∩ C) / P(C) = 0
P (C | A) = P(C ∩ A) / P(A) = P(A ∩ C) / P(A) = 0
P (B | C) = P (B ∩ C) / P(C) = 2/11
(b)
P(A ∩ B) = 6/36 = 1/6 is not equal to 0, hence A and B are not disjoint.
Also P(A)*P(B) = (15*10)/(36*36) is not equal to P(A ∩ B) = 1/6, hence A and B are not independent.
So, the events A and B are neither independent nor disjoint.