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one or more "7"

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by sanju09 » Tue Feb 24, 2009 4:32 am
What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits?

A. 271/900

B. 27/100

C. 7/25

D. 1/9

E. 1/10
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by sureshbala » Tue Feb 24, 2009 5:01 am
Total number of 3 digit numbers = 9x10x10 = 900

Now let us calculate the numbers that do not contain a 7.

The hundredth's place of this number can be filled in 8 ways where as the remaining two digits can be filled in 9 ways each.

So the total number of 3 digit numbers that do not contain digit 7 = 8x9x9=648.

Hence the total number of 3 digit number that will contain at least one 7 = 900-648 = 252.

So required probability = 252/900 = 7/25
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