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

# A 20 b 25 c 100 d 150 e 200 using the probability

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Unformatted text preview: heir tens digit is even? A) 20 B) 25 C) 100 D) 150 E) 200 Using the probability method N = 600-200 = 400 P = 5/10*5/10 = 25/100 5/10 is for the even tens and 5/10 is for the last digit being odd (thus) making the integer odd. N*P = 25/100*400 = 100 The answer is (C) How many 4-digit positive integers are there in which all 4 digits are even? A) 625 B) 600 C) 500 D) 400 E) 256 Using the probability method N = 10,000-1,000 = 9,000 P = 4/9*5/10*5/10*5/10 = 500/9000 = 5/90 N*P = 9,000*5/90 = 500 The answer is (C) Since the first digit cannot be zero (i.e. 01111 is not a four digit number) thn the probability that the first digit will be even is 4/9. For the other three digits we can choose from 0 to 9 so 5/10. 2 How many 7 digit numbers are even and have a "3" in the hundreds place? Using the probability method N = 10,000,000 – 1,000,000 = 9,000,000 P = 1/10*5/10 = 5/100 N*P = 9,000,000*1/20 = 450,000 Since the number is even it has to end with an even digit, so 5/10 for the last digit, the hundreds place has to be 3 s...
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