HW 14 Equally Likely outcomes Conditional Probability and In
Solution
1.
For each character, there are 26+26 + 10 = 62 possibilities.
As they can repeat, then there are 62^6 = 56800235584 OUTCOMES in sample S. [answer]
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2.
For each character, there are 26+26 = 52 possibilities.
As they can repeat, then there are 52^6 = 19770609664
OUTCOMES in event A. [answer]
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3.
For each character, there are 10 possibilities.
As they can repeat, then there are 10^6 = 1000000
OUTCOMES in event B. [answer]
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4.
ZERO. Events A and B do not intersect as event A is all letters, while B is all numbers.
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5.
P(A) = 19770609664/56800235584 = 0.34807267 [answer]
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6.
P(A|B) = P(A and B) / P(B)
as P(A and B) = 0 [mutually exclusive]
P(A|B) = 0 [answer]
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7.
No, as P(A) =/ P(A|B).
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