Probabilities: Problems with Solutions

Solution:
The sample space has [tex]6 \times 6 \times 6 = 6^3 = 216[/tex] equally likely outcomes. Only one of them, [tex](6,6,6)[/tex], gives three sixes. If [tex]A[/tex] is the event “three sixes”, then
[tex]P(A) = \frac{1}{6^3} = \frac{1}{216}[/tex]

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 is only one outcome, [tex](1,1)[/tex], and let C be the event "sum equal to 3". There are two outcomes, [tex](1,2)[/tex] and [tex](2,1)[/tex]. So:
[tex]P(A \cup B \cup C) = P(A) + P(B) + P(C) = \frac{0}{36} + \frac{1}{36} + \frac{2}{36} = \frac{3}{36} = \frac{1}{12}[/tex]

Solution:
There are 4 queens in 52 cards, so the probability is [tex]\frac{4}{52}=\frac{1}{13}[/tex].

Solution:
It is clear that P(a) + P(b) = 1, on the other hand [tex]\frac{P(a)}{P(b)}=\frac{5}{11} \Rightarrow P(a) = P(b) \times \frac{5}{11}[/tex].
[tex]P(a) + P(b) = 1 \Rightarrow P(b) \times \frac{5}{11} + P(b) = 1[/tex]
[tex]P(b) \times \frac{16}{11} = 1[/tex]
[tex]P(b) = \frac{11}{16}[/tex]

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 [tex]P(A) = \frac{9}{36}=\frac{1}{4}[/tex]

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