Problem Solving

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



Answer given is C

My approach was to find the probabilty that 7 doesnt appear in the 3 digit no. i.e for each digit 9 nos can be chosen out of 10 (excluding 7) so the probabilty is 9x9x9/10x10x10. i.e 729/1000.
Therfore the anwer should be 1-729/1000 which is 271/1000. since its not available in the option i took the nearest value which is 27/100. Please explain what am i doing wrong here. Thanks.

No of 3 digit integers are not 1000 but 900 rest being one digit or 2 digit
Since it is 3 digit no therefore 0 cannot occupy hundred place. therefore no of cases are 8*9*9=648(without 7 as a possible digit)

now 900-648=252 no of 3 digits integers that have exactly one 7
reqd probability=252/900=84/300=28/100=7/25

Answer should be C

rah_pandey wrote:No of 3 digit integers are not 1000 but 900 rest being one digit or 2 digit
Since it is 3 digit no therefore 0 cannot occupy hundred place. therefore no of cases are 8*9*9=648(without 7 as a possible digit)

now 900-648=252 no of 3 digits integers that have exactly one 7
reqd probability=252/900=84/300=28/100=7/25

Answer should be C

Thanks i feel real stupid not considering the obvious.

Hey Rah_pandey

Hey rah_pandey

"No of 3 digit integers are not 1000 but 900 rest being one digit or 2 digit"

why didnt you take 999 instead of 900???

Total no of integers from 0-99 =100
100-999=900 therefore 900 3 digit positive integers

I see how you guys did the problem, but can someone explain why it would be incorrect to do the problem like this:

probability that 7 is in the hundreds digit = 1/9

prob that 7 is in the tens digit= 1/10

prob that 7 is in the units digit = 1/10

1/9 + 1/10 + 1/10 = 29/90

I know its not an answer choice but why is that approach wrong?

osirus0830 wrote:I see how you guys did the problem, but can someone explain why it would be incorrect to do the problem like this:

probability that 7 is in the hundreds digit = 1/9

prob that 7 is in the tens digit= 1/10

prob that 7 is in the units digit = 1/10

1/9 + 1/10 + 1/10 = 29/90

I know its not an answer choice but why is that approach wrong?

osirus,
how about probability that 7 is in the hundreds digit and 7 is in the tens digit,
then you caculate probability that 7 is in the tens digit and 7 is in the unit's digit, and so on....
you see where you are going..

but you could solve this like this too if it makes things easier for you:

3 digit numbers containing 7s:
X07, X17, X27... X97; X= 1,2,3,4,5,6,8,9 - total will be 10*8=80
X70, X71, X72... X76, X78, X79; total will be 9*8=72
700, 701... 199 - total of 100

total=252
prob=252/900

Hope I helped