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

A 24 b 36 c 72 d 144 e 216 using the probability

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Unformatted text preview: ning digit different from the other two? A) 24 B) 36 C) 72 D) 144 E) 216 Using the probability method N = 1,000 – 100 = 900 P1 = 1/10*8/10 = 8/100 P2 = 8/10*1/10 = 8/100 P3 = 8/10*1/10 = 8/100 SP = 8/100+8/100+8/100 = 24/100 N*P = 900*24/1,000 = 216 The answer is (E) 4 For calculating P we need this time to understand that again there are three possible P's - but with a twist. In P1 the first digit is not important, let's say we choose 1. The second digit has to be also 1 but we have 10 digits to choose from (0-9) so the probability is 1/10 and the last digit has to be different than 1 but we cannot choose zero so 8/10. In P2 the first digit is not important (let's say we choose 8) but the second digit can't be 8 and can't be zero so 8/10 and the last digit has to be 8 so 1/10. In P3 the same logic is used. The first digit is not important (let's say we choose 5) – in the second digit we can choose 8 out of 10 digits (we can't choose 5 which we've already chosen, nor can we choose 0) so – 8/10 and in the last digit we need to choose the second digit exactly so 1/10. Written by KillerSquirrel – 2007 5...
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This document was uploaded on 02/28/2014.

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