Probabilities: Problems with Solutions

Solution:
The sample space has 6 ⋅ 6 ⋅ 6 = 6³ members. One of the member of the sample space is (6,6,6), so if we denote the event of having three 6 by A then: P(A) = 1/6³

Solution:
All possibilities are:
(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)
Let A be the event "sum equal to 1". There are no outcomes which correspond to a sum equal to 1.
Let B be the event "sum equal to 2". There are only one member, which is (1,1), and let C be the event "sum equal to 3". There are two members, which are (1,2) and (2,1). So the last answer is: P({A,B,C}) = P(A) + P(B) + P(C) = 0/36 + 1/36 + 2/36 = 3/36 = 1/12

Solution:
There are 4 queens in 52 card so the probability is 4/52=1/13.

Solution:
It is clear that P(a) + P(b) = 1, on the other hand P(a)/P(b)=5/11 => P(a) = P(b)*5/11.
P(a) + P(b) = 1 => P(b)*5/11 + P(b) = 1
P(b)*16/11 = 1
P(b) = 11/16

Solution:
All possibilities are:
(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)
Let A be the event "sum is multiply of 4". A contains (2,2), (1,3), (3,1), (2,6), (6,2), (3,5), (5,3), (4,4), (6,6); so P(A) = 9/36=1/4

Solution:
[tex]P(A) = \frac{{{10}\choose{2}}}{1} = \frac{45}{1024}[/tex]

Solution:
P(A∪B) = P(A) + P(B) - P(A∩B') =>
0.8=0.4 - P(A∩B) =>P(A∩B)=0.2
P(A∩B') = P(A) - P(A∩B) = 0.4 - 0.2 = 0.2