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Unformatted text preview: not important, let's say we choose 7. The second digit has to be also 7 so the probability is 1/10 and the
last digit has to be different than 7 so 9/10.
In P2 the first digit is not important (let's say we choose 8) but the second digit can't be 8 so 9/10 and the
last digit has to be 8 so 1/10.
In P3 the same logic is used. The first digit is not important – in the second digit we can choose nine out of
ten digits (the only one we can't choose is the first digit) so – 9/10 and in the last digit we need to choose
exactly the second digit so 1/10. We can think about it like this:
For P1 = 77x or 88x or 99x
For P2 = 7x7 or 8x8 or 9x9
For P3 = 7xx or 8xx or 9xx
The answer is (C). 3 How many times will the digit 7 be written when listing the integers from 1 to 1000?
This is the problem we started with (see above) - Using the probability method to solve:...
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- Fall '13