What is the probability that a 3-digit positive integer picked at rand

Question

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

I guessed A, pressed for time. How would I do this problem without manually figuring out how many digits have at least one 7?

When I did it again for correctness, I did the following:

|{700...799}| = 100 integers.
|{107, 117, 127, 137, 147, 157, 167, 187, 197}| = 9 per set of 100, therefore, 72 integers (8 sets, excluding 700s).
|{170, 171, 172, 173, 174, 175, 176, 178, 179}| = 9 per set of 100, therefore, 72 integers (8 sets, excluding 700s).
|{177, 277, 377, 477, 577, 677, 877, 977}| = 8 integers

Total integers with at least one 7 = 100 + 72 + 72 + 8 = 252
Total outcomes = 999 - 100 + 1 = 900

Probability = 252/900 = 7/25. That. took. forever. :evil:

Bunuel · Accepted Answer

There are total 900 3 digit numbers;

3 digit number with no 7 =8*9*9=648 (first digit can take 8 values from 1 to 9 excluding 7; second and third digits can take 9 values from 0 to 9 excluding 7),

P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25

Answer: C.

jeeteshsingh · Answer

Question  a 3-digit positive integer picked at random will have one or more "7" in its digits 
can be approached by finding out the number of positive integers which would not have 7 at all and then subtracting the posb from the overall posb combinations.

3 Digit Number posb = 9 x 10 x 10 = 900

3 digit number not having 7 in it all = 8 x 9 x 9 = 648

Therefore 3 digits number having atleast one 7 in them = 900 - 648 = 252

Hence probability  = \frac{252}{900} = \frac{7}{25} . Hence answer is  C

Joy111 · Answer

some one please point the flaw in my logic

want to do it the longer way for more understanding .

at least one 7 means , one , two or three 7's

lets take scenarios, first digit 7 , two digits 7 , all three digits 7

1) first digit to be seven , non seven , non seven

(1/9 ) * 9/10 * 9/10 = 81/900=9/100 first digit cannot be zero , so there are 9 total possibilities for the first , from

1 ,2 , 3, 4, 5, 6, 7, 8, 9. For the second and third we have 10 possibilities now we can have zero , from which favorable are nine, , as we are are excluding 7

( case 2) 7 , 7 , non seven

1/9 * 1/10 *9/10 * 3 = 3/100 here we are multiplying by 3 , because the 2 seven's can occur in 3 ways

( case 3) 7, 7 , 7

1/9 * 1/10 * 1/10 = 1/ 900

so at least one 7

9/100 + 3/100 + 1/900 = 109/900 = wrong answer !! :oops:

what am I missing guys ?

I know the alternate way , 1- p ( all not seven ) , that's fine , but I want to understand what's wrong with this way .

Appreciate your help. Thank you

what am I missing guys ? I know the alternate way , 1- p ( all not seven ) , that's fine , but I want to understand what's wrong with this way . Appreciate your help. Thank you

gmatsaga · Answer

Oh man my mistake was 9 * 9 * 9 and then you mentioned that first digit can only take values from 1 2 3 4 5 6 7 8 9 and not 0 because that will make it a 2 digit number.

Argh! traps traps traps!

Only1 · Answer

You are considering 'only one 7' as your first case by counting numbers in the form of 7xy,7xx,7yy. What about x7y,y7x,xx7,yy7,xy7,yx7?? By counting such possible numbers, the number of events will increase. Hope this helps.

Nikhil30 · Answer

Bunuel  can you tell me where i went wrong.
atleast one seven means : one,two or three 7:
p= no. fav.outcome/ total no of outcome.
total no of outcome = 900
now 100-199= 20's (seven)
200-299= 20's seven 
there for 20*9= 180's seven .
excluding 700-799 = no of seven is 20+100=120's seven.
therefore no of seven = 180+120=300's seven.

therefor p= 300/900 which is equal to 1/3 ,

please tell me where i went wrong.

Bambi2021 · Answer

From 100-200 there is 10 (tens digit) + 9 (units digit except for 177 which is already counted)
From 200-300 there is 10 (tens digit) + 9 (units digit except for 277 which is already counted)

19*9 = 171

From 700-800 there is another 100-19 = 81 left

171+81 = 252

252/900 is, well, here is the hardest part because I was not able to boil this down to 7/25, but the exclusion method worked pretty well from here.

Nikhil30 · Answer

From 100-200 there is 10 (tens digit) + 9 (units digit except for 177 which is already counted)
From 200-300 there is 10 (tens digit) + 9 (units digit except for 277 which is already counted)

19*9 = 171

From 700-800 there is another 100-19 = 81 left

171+81 = 252

252/900 is, well, here is the hardest part because I was not able to boil this down to 7/25, but the exclusion method worked pretty well from here.

hey , between 100-199 = there are 20's seven unit digit : 10's seven and tens digit : i.e 170,171,172,173,174,175,176,177,178,179 : equal to 10's seven there for 
between 100-199= total no of seven = 10+10=20 .
am i right ??

Bambi2021 · Answer

You are right that there are 10 tens digits and 10 units digits, but dont forget that you have the number 177 where both digits is a 7, so you dont want to count this number twice. Therefore 20-1 = 19 unique numbers that contain a 7.

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What is the probability that a 3-digit positive integer picked at rand

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What is the probability that a 3-digit positive integer picked at rand

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