thephoenix wrote: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

There are 900 ways to pick up a random 3 digit positive number

Now no of ways in which we can select number which does not contain 7 = 8*9*9 = 648

No of ways in which we can select a number which contains 7 = 900 -648 = 252

so the required probability = 252/900 = 28/100 = 7/25

Hence C

What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits?


Total Number of 3 digit numbers are 900 (i.e) 100 to 999 including.

p(one or more 7's in the 3 digits) = 1 - p(no 7 in the 3 digits)

Lets say XYZ is number with out digits 7 then

X can be digits 1, 2 ,3 4, 5, 6, 8, 9 = 8 ways
Y can be digits 0, 1, 2 ,3 4, 5, 6, 8, 9 = 9 ways
Z can be digits 0, 1, 2 ,3 4, 5, 6, 8, 9 = 9 ways

p(one or more 7's in the 3 digits) = 1- p(no 7 in the 3 digits)/ Total number of ways.

p(one or more 7's in the 3 digits) = 1- 8*9*9/900 = 1-648/900 = 252/900 = 7/25

So, answers is 'C'