An integer x, with 10

Question

An integer x, with  10\leq x\leq 99 , is to be chosen. If all choices are equally likely, what is the probability that at least one digit of x is a 7?

(A) 1/9
(B) 1/5
(C) 19/90
(D) 2/9
(E) 1/3

Archit3110 · Answer

total integerts = 99-10 +1 ; 90
so with 7 ; 19 integers possible from 10 to 99
P = 19/90
IMO C

@bunuel; is given answer correct? if yes then please provide solution

SWAPNILP · Answer

I got answer 1/5. But i am not sure how it is right. Request Bunnel to explain.. Thanks

Abhigmat2019 · Answer

optn B... 18/90..... OA pls...

Sent from my iPhone using GMAT Club Forum

Archit3110 · Answer

Abhigmat2019 
77 has 2 7's so it will be 19 7s...

Archit3110 · Answer

SWAPNILP 
please share your solution of how you got 1/5.

Riya06 · Answer

I initially picked C, but the problem is we are taking '77' twice while getting to 19. Therefore, it should be 18/90 = 1/5

Option B

SWAPNILP · Answer

Hi Bunuel

Please find below my approach:
Step 1: total no of integers between 10 and 99 = 90
Step 2: Integers with digit 7 = 18
Step 3: Probability = 18C1/90C1 = 18/90 = 1/5

Please advise.

Shrey9 · Answer

We have 99-10 = 89+1 (inclusive set) = 90 integers total..

We are asked AT LEAST ONE DIGIT OF X TO BE A 7 i.e we are okay even if we get a 77 and it won't be double counting because
 "At least one 7 means", 2 or 3 or 4 or 5 or , as many 7's as possible are welcome in this combination.. but since the upper limit is 99, we have to stop at 97 here, because that's the last digit with AT LEAST ONE 7 In it.

therefore, we start with first digit with a 7 in it and go on...
 17, 27, 37, 47, 57, 67, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 87, 97 ..

We get 18 integers with AT least one 7..
and when divided by 90, which is the total number of integers in the given list.

we get 
18/90 = 1/5

(B) as the answer.

dave13 · Answer

HOW TO FIND NUMBER OF TERMS FROM A TO Z \frac{last..term - first..term}{2} +1 First lets find number of terms between 10 and 99 inclusive \frac{99 - 10}{2} +1 = 45 we can have following numbers with at least one digit 7. And why everyone loves this number 7

17, 27, 37, 47, 57, 67, 77, 87, 97 So total 9 numbers Hence the probability that at least one digit of x is a 7 = \frac{9}{45} or \frac{1}{5}

Archit3110 · Answer

dave13 
any reason why you havent considered no from 70 to 79...?

dave13 · Answer

Archit3110 i actually used a shortcut, cause its less time consuming, well i think so

i used these numbers: 17, 27, 37, 47, 57, 67, 77, 87, 97 but if i reverse these numbers i get: 71, 72, 73, 74, 75, 76, 78, 79 and also 70 So total numbers now if you see 18: \frac{18}{90} = \frac{1}{5} same result

So if I consider, 70 to 79, then I have to consider the rest numbers that I have reversed

Archit3110 · Answer

dave13 ok, understood, but i dnt think this trick would be always helpful.. anyways thanks for enlightening on an alternate method..

geezus24x7 · Answer

If anyone's interested in a solution without counting all possible number, here's one-

Number of ways a number can be picked = (99-10) + 1=> 90
An easier way to calculate probability to pick a number with atleast one 7 is: 1 - (Probability of picking a number with no 7s)

Probability of picking a number with no 7s-
Since the number has to be a two digit number, number of ways in which its ten's digit can be selected is 8 (since we can't use 7 or 0)
 AND , number of ways in which its ten's digit can be selected is 9 (since we can't use 7)
Therefore, Probability of picking a number with no 7s = 72/90 or 4/5

And hence probability to pick a number with atleast one 7= 1 - 4/5 = 1/5

Ghoskrp3 · Answer

total integers = 99-10 +1  => 90  (satisfying 10≤x≤99)

So with 7 either in units or tens place;  18 integers  are possible from 10 to 99 inclusive 
 (as 17,27,37,47,57,67,87,97 -> 8 integers & 70,71 ... 79 -> 10 integers: making it a total of 18 integers)

Probability = 18/90 => 1/5

(B) 1/5

An integer x, with 10 <= x <= 99, is to be chosen. If all choices are

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