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

# One footnote on the concept of xx i said that you can

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Unformatted text preview: the probability method. One footnote on the concept of XX: I said that you can choose out of ten digits for the first X and out of ten digits for the other X. This is not entirely true since 00 is not a valid integer (out of range) but 100 is also out of range (XXX) so these two integers cancel out each other (you will see that this will give us a headache in more complex problems). 1 How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3? A) 10,300 B) 10,030 C) 1,353 D) 1,352 E) 1,339 Using the probability method we need to find the number of integers (N) so here N = 458,600 – 324,700 = 133,900 and now we need to figure out the probability (P). Since we have two digits XX for the tens digit and units digit and the probability for the first X is 1/10 and the probability for unit digit of 3 is 1/10 then P = 1/10*1/10 = 1/100. P*N = 1/100*133,900 = 1,339 And the answer is (E) How many odd integers between 200 and 600 are there such that t...
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