Solve for z -cz+6z=tz+83
1 answer:
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Answer and explanation:
Given : Suppose you and a friend each choose at random an integer between 1 and 8, inclusive.
The sample space is
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (1,7) (1,8)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (2,7) (2,8)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (3,7) (3,8)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (4,7) (4,8)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (5,7) (5,8)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) (6,7) (6,8)
(7,1) (7,2) (7,3) (7,4) (7,5) (7,6) (7,7) (7,8)
(8,1) (8,2) (8,3) (8,4) (8,5) (8,6) (8,7) (8,8)
Total number of outcome = 64
To find : The following probabilities ?
Solution :
The probability is given by,
a) p(you pick 5 and your friend picks 8)
The favorable outcome is (5,8)= 1
b) p(sum of the two numbers picked is < 4)
The favorable outcome is (1,1), (1,2), (2,1)= 3
c) p(both numbers match)
The favorable outcome is (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7), (8,8) = 8