Without doubt D

See here more on probability https://gre.myprepclub.com/forum/gre-tips- ... 23084.html

Consider these two cases.....

Case i: Each apartment has one occupant, EXCEPT for apartment #4, which has 1 BILLION occupants.
In other words, 1,000,000,083 people live in the apartment, 1,000,000,000 of whom live in apartment #4.
So if we randomly select someone from the apartment, it's very very very very very likely that that person lives in apartment #4.
So, in this case P(apartment number of a randomly selected tenant contains a 4) = a number very close to 1.
In this case, QUANTITY A IS GREATER

Case ii: Each apartment has one occupant, EXCEPT for apartment #5, which has 1 BILLION occupants.
In other words, 1,000,000,083 people live in the apartment, 1,000,000,000 of whom live in apartment #5.
So if we randomly select someone from the apartment, it's very very very very very likely that that person lives in apartment #5.
So, in this case P(apartment number of a randomly selected tenant contains a 4) = a number very close to 0.
In this case, QUANTITY B IS GREATER

Answer: D

such questions are discriminatory in nature.

The answer is C

The total number of apartment numbers containing 4 is 18. Hence 18/(85-2+1)=18/84 = 3/14. Hence C.

This solution says the answer is C

Step 1: Total Apartments
The apartments are numbered 2 through 85, so the total number of apartments is:

85−2+1=84.
Step 2: Apartments Containing the Digit 4
We need to count how many apartment numbers contain the digit
4
4. These numbers can be in either the tens place or the units place.

Case 1: Apartments with 4 in the Tens Place
The apartments with

4 in the tens place are:

40,41,42,43,44,45,46,47,48,49
There are
10
10 such apartments.

Case 2: Apartments with 4 in the Units Place
The apartments with
4 in the units place are:

4,14,24,34,44,54,64,74,84
There are
9 such apartments.

Overlap: Apartment 44
Apartment
44
44 is counted in both cases, so we subtract it once to avoid double-counting:

10+9−1=18 apartments contain the digit 4
.
10+9−1=18 apartments contain the digit 4.
Step 3: Probability of Selecting an Apartment with the Digit 4
The probability is:

P(Apartment contains a 4)
=
Number of favorable outcomes
Total outcomes
=
18/84
=
3/14
.
P(Apartment contains a 4)=
Total outcomes
Number of favorable outcomes
​
=
84/18
​
=
14/3
​
.

​
.
The two quantities are equal.

The solution doesn't even make sense. It has to be C. If we have to nitpick, every question could be incorrect.

The language of the question is quite ambiguous. I mean since they haven't mentioned the number of tenants, it could be D.
Otherwise it's clear cut - C

damn, the trap is in the wording, theres no limit, details about how many tenants per apartment so lets assume that each appartment has 1 tenant and the result is 9/84
ok, next what if only 1/3 of the apartments are occupied then its 3/84, so its D. I got tricked the first time thinking the question is asking for probability of picking a random apartment but its actually picking a random tenant.