Unformatted text preview: o 1/10 is the probability for that.
Of the three-digit integers greater than 700, how many have two digits that are equal to each
other and the remaining digit different from the other two?
Using the probability method
N = 1,000 – 700 = 300
P1 = 1/10*9/10 = 9/100
P2 = 9/10*1/10 = 9/100
P3 = 9/10*1/10 = 9/100 SP = 9/100+9/100+9/100 = 27/100
N*P = 300*27/100 = 81
As you can see in the given answers, there isn't such an answer 81. This is due to the fact that the answer
asks us for numbers that are greater then 700 and less then 1,000. If we let N = 300 this means 700 and
1,000 are in the range. If you recall I warned you this will sometimes give us a headache. We need to
remove 700 from N*P so the answer is N*P-1 = 81-1 = 80.
For calculating P we need to understand that there are three possible P's. In the first one (P1) the first digit
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- Fall '13
- Numerical digit, digit, 4-digit, 4 digits, probability method