GMAT Club Olympics 2025 (Day 1): Ann and Bob each randomly select an

I got this question wrong, but actually, the first option is sufficient.
We can actually find favourable outcomes for Ann from the first option, since from the first option,we now know there are ten consecutive numbers.
And if we do 10C2 we get combinations of two numbers chosen from the list.
For example, in 1,2,3,4,5,6,7,8,9,10.The combination of {3,7} where ann chooses 3,she loses,but if you reverse its
{7,3} Ann wins.
So for each combination of numbers, Ann wins.
So the total favourable outcomes is 10C2.

However in the second option there is no range,and hence not sufficent.

1) y - x = 9
This tells us the range has 10 integers, but we don’t know the actual values of x and y.
The probability depends only on the number of possible values, not which ones.
We can compute the exact probability from this info.
A is sufficient, (it could be A or D at this point)

2) this just gives upper limit, so we have nothing on probability
This is insufficient, hence A would be the answer.

A. Because (2) gives y alone is not sufficient. (1) y-x=9 -> Probability = y-x/2(y-x+1) = 9/20

Ann and Bob each randomly select an integer from the integers x to y, inclusive. What is the probability that the integer Ann selects is greater than the one Bob selects?

The probability depends on whether x & y are equal or not. If they are equal probability becomes 0, if not 1 would be greater than anyways.
St 1 tells us that both are unequal and hence sufficient.
S2 only tells about y, so x may or may not be equal to, hence insufficient

otal number of ordered pairs (Ann’s pick, Bob’s pick) = n · n = n2.
Number of those pairs with Ann’s pick > Bob’s pick = the number of 2-element subsets of the n values, since each unordered pair {a, b} with a ≠ b yields exactly one ordered pair (a, b) with a > b. That count is C(n, 2) = n(n – 1)/2.
Therefore
P(Ann > Bob) = [n(n – 1)/2] / n2 = (n – 1)/(2n).

Statement (1): y – x = 9 ⟹ n = 10 ⟹ P = (10 – 1)/(2·10) = 9/20. ⇒ sufficient.
Statement (2): y = –20 fixes only y, leaves x (and hence n) unknown. ⇒ not sufficient.

Ann and Bob each randomly select an integer from the integers x to y, inclusive.

What is the probability that the integer Ann selects is greater than the one Bob selects?

(1) y - x = 9; y = x+9
Integers = {x, x+1, x+2, x+3, x+4, x+5, x+6, x+7, x+8, x+9}
Total cases = 10*10 = 100
Let the number of cases when Ann selects is greater than the one Bob selects = The number of cases when Bob selects is greater than the one Ann selects = x
The number of cases when both select the same integers = 10
2x + 10 = 100
x = 45
The probability that the integer Ann selects is greater than the one Bob selects = 45/100
SUFFICIENT

(2) y = -20
Since x is unknown
NOT SUFFICIENT

IMO A

Let the set of possible integers be S=x,x+1,...,y S = {x, x+1, ..., y} S=x,x+1,...,y.

The total number of possible choices for each person: n=y−x+1
Total possible pairs: n^2
Number of pairs where Ann's integer is greater than Bob's: For each possible value Ann can pick, count how many values Bob can pick that are less.

Let’s formalize:
For Ann to pick a number greater than Bob, Ann can pick any number from x+1 to y , and for each such number k, Bob can pick any number from x to k-1.

So the number of favorable pairs can be given as - number of favorable pairs/n^2 = (y-x)*(y-x+1)/2/(y-x+1)^2

Since we know the value of y-x=--9, we can formulate our answer for this question , hence 1 is sufficient.

For statement 2,
y=-20. We do not know x . The range could be any length, so the probability cannot be determined with this statement alone, hence we can rule out options (B) & (D).

Hence the answer is option (A)

Ann and Bob are selecting independently from the same set of integers: from x to y, inclusive.

The total number of integers =n=y−x+1

Let’s call the integers:x,x+1,x+2,...,y

Ann and Bob each pick one. We want to know the probability that Ann’s number is greater than Bob’s.

Note: If we know how many distinct integers are in the set, we can compute all possible outcomes and count the favorable ones (Ann > Bob).

🟩 Statement (1): y − x = 9
Then:

Number of integers = 9 + 1 = 10

So the set has 10 integers.

Since both are choosing uniformly from the same set of 10 integers, and we know the number of values, we can:

Calculate total outcomes: 10 × 10 = 100 pairs

Count favorable outcomes (Ann > Bob)

We don’t need to know x or y individually, only the number of values, which we have.

✅ Sufficient

🟩 Statement (2): y = -20
This tells us y, but nothing about x.

So we don’t know the size of the set of integers from x to y. Could be 1 number, 5 numbers, 100 numbers...

Without knowing x, we can’t compute how many integers are in the set → we cannot find the total number of possible outcomes.

🚫 Not sufficient

🟩 Combined:
(1) is already sufficient on its own, so we don’t need to combine.

✅ Final Answer: (A) — Statement (1) alone is sufficient, but statement (2) alone is not.

What is the probability that the integer Ann selects is greater than the one Bob selects?

S1
y - x = 9
This means there are a total of 10 integers to choose from
Now, total number of selections = 10*10 = 100
Out of these 100 scenarios, there will be 10 scenarios when both numbers are equal
Of the remaining 90 scenarios, owing to symmetry, there are exactly half of the scenarios when the values are greater for 90
So, P = 45/100 =.45
Sufficient

S2
y = -20
No info on x and hence, the range
Insufficient

Answer A

(2) y = -20

Knowing one of the numbers is not sufficient. So, the easiest to eliminate is B and D.

(1) y - x = 9

The information tells us that there are 10 numbers in the series. Knowing this information is sufficient as we can find the probability.

Out of the number of subsets that can be created, half will have Ann > Bob

10C2/2*10 = 9/20

Sufficient

Option A

I couldnt directly click on the correct option so my reply is E.

First, let's understand what we're looking for. If Ann and Bob are each randomly selecting integers from the same range (x to y inclusive), we need to determine the probability that Ann's selection is greater than Bob's.
Statement 1: y - x= 9
This tells us the range has 10 integers (from x to y inclusive). However, we don't know the actual values of x and y. Let's see if we can determine the probability anyway.
For any two randomly selected integers from this range, there are a total of 10 × 10 = 100 possible outcomes (each person has 10 options). For Ann to select a greater number than Bob, we need to count the favorable outcomes.
If the range is [x, x+1, x+2, .., y), then for each value Bob selects, Ann needs to select a larger value. For example, if Bob selects x, Ann can select any of the 9 values greater than x. If Bob selects x+1, Ann can select any of the 8 values greater than x+1, and so on.
The number of favorable outcomes is: 9 + 8 + 7+6+5+4+3+2+1+0=45
So the probability is 45/100 = 9/20
Statement 1 alone is sufficient.
Statement 2: y = -20
This only tells us the upper bound of the range. Without knowing x, we can't determine how many integers are in the range, so we can't calculate the probability.
Statement 2 alone is not sufficient.
The answer is A) Statement 1 alone is sufficient, but statement 2 alone is not.

The answer is A.

The total number of integers they can choose from in this case would be y-x+1. (Eg: If I'm choosing numbers from 1 to 5 inclusive, I would do 5-1+1 which is 5. the numbers would be 1,2,3,4,5)

In this case because each of them randomly select an integer from x to y, the probability of them both choosing any integer would be the same. We also know that there would be some cases where they would choose the same number and it would be a tie. For eg: If they could choose any number from 1 to 3 - there would 3 cases where they could both choose the same number (1,1), (2,2),(3,3)

Since the probability of either of them getting an integer more than the other is the same, we can calculate the probability in which one would get more by

P = favourable outcomes / total outcomes

Favourable outcomes = n^2 - n where n^2 are the total number of possibilities and n are the number of ties as explained above.
Total outcomes = n^2 (for eg: in the example above, there would be 3 * 3 possibilities of choosing numbers)


So P = n^2-n/ n^2
Therefore, P = (y-x+1)^2 - (y-x+1)/ (y-x+1)^2.

Looking at options, in statement I: we see y-x is given to us so y-x+1 would be 10. We can therefore calculate the probability.
Statement II gives us the value of y but we know nothing about x so calculating probability is not possible.

Statement (1) With the range, given in the statement, we can calculate the probability for Ann for each possible choice of Bob. Example: if Bob pick the lowest number, Ann will have 8/9 chance. If Bob pick the second lowest, Ann will have 7/9 and so on. Since it's a DS question, I will stop here. Eliminate choices B, C, and E.

Statement (2) With the value of y we can not have any useful information to calculate the probabilities. Eliminate choice D.

Answer = A.

Now from 2nd statement substituting the values we will get x also so we get the range of integers which is x=-29 to y=-20. Now we can easily find the probablity of Ann selecting greater so answer would be Can be determined from both statements but not single statement alone.

Let the number selected by Bob be n.

And n lies in the range x <= n <= y, where x,y are both integers.

We need to find : Probability that Integer selected by Ann is greater than the one Bob selects.

Statement 1:

(1) y - x = 9

y-x = 9, Hence, y = 9+x

x<= n <= 9+x

if x = 1, then y = 9+x = 9+1 =10

1 <= n <= 10.

if Bob selects, 1, Ann has 9 options 2,3,4,5,6,7,8,9,10.

if Bob selects 2, Ann has 8 options 3,4,5,6,7,8,9,10.

Probability of Ann selecting = 9(9+1)/2 = 45

Total probability of Ann = 45/100

Probability of Ann > Bob = Probabilty of Bob > Ann = 0.45

probability of both selecting same = 1-(0.45+0.45) = 0.10

Hence, Statement 1 is sufficient

Statement 2:

(2) y = -20

x can assume a wide array of values. Without any concrete value of x, we cannot find the probabilities.

Hence, insufficient

Option A

Answer: 9/10
Y= -20 and X= -29 (solving equation we get)
Now Anna and Bob can select from [-20,-29]
So Total we have 10 numbers and 9 cases can be made,
Answer is 9/10