Question

In: Statistics and Probability

The following events are considered for a throw with two dice: A: The total of eyes...

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?

Solutions

Expert Solution

(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.

orchestra answered 1 year ago
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