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A computer generates a two digit number random number. It can be any number from 00 to 99.
Find the probability that it.
(A) - is 99
(B) - is not 99
(C) - has no 9s in it
(D) - had at least one 9 in it.
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Notice that in all, there are 100 two-digit numbers from 00 to 99.
(A) The answer to (A) is obvious: it is P = = 0.01.
(B) The answer to (B) is the complement to value of (A): P = 1 - = = 0.99.
(D) There are 100 two-digit numbers, and each such number has two digits.
So, the count of the digits is 2*100 = 200: two hundred digits are in use.
It is obvious, that all the digits have equal rights and, therefore,
you can meet them equally often. It means that the digit "9" is used 200/10 = 20 times.
In turn, it means, that in all, there are 19 two-digit numbers, where "9"
does present (not 20, but 19 numbers, because in the number 99 the digit "9"
is included twice).
THEREFORE, the answer to (D) is P = = 0.19.
(C) The answer to (C) is the complement to value of (D): P = 1 - = = 0.81.
Solved.
I answered all the questions.
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When I saw the number 0.81 as the answer in (C), it came to my mind
that the solution for (C) could be constructed in different (simpler) way.
Indeed, there are 9 digits from 0 to 9, excluding "9": {0, 1, 2, 3, 4, 5, 6, 7, 8}.
Hence, there are 9*9 = 81 two-digit numbers, where "9" does not present.
It gives the probability = 0.81 in (C) by easy way.
Solved twice, by two different methods, for your better understanding.