A%20Probability%20Approach%20For%20Solving%20Counting%20Problems

The second digit has to be also 7 so the probability

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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? A) 110 B) 111 C) 271 D) 300 E) 304 This is the problem we started with (see above) - Using the probability method to solve:...
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