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11 Jun 2026, 23:45
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total dogs= 9
num of lab = x
others = 9-x
probability of selecting two labs = (x/9) * (x-1)/8 = p
p<1/2
(x/9) * (x-1/ 8) < 1/2
x * (x-1)< 36
we need to find value of x.
S-1
P(lab) * P(non lab) + P(non lab)* P(lab) > 1/2
(x/9) * (9-x/8) + (9-x/8) * (x/9) > 1/2
(9-x)x > 18
for this to be true,
x must be less than 6.
so this is sufficient to answer.
S-2
(9-x/9) * (8-x/8) >1/10
(9-x)(8-x) > 7.2
for this to be true, x must be less than 6.
so this is sufficient as well.
each option alone is sufficient.
Option D
num of lab = x
others = 9-x
probability of selecting two labs = (x/9) * (x-1)/8 = p
p<1/2
(x/9) * (x-1/ 8) < 1/2
x * (x-1)< 36
we need to find value of x.
S-1
P(lab) * P(non lab) + P(non lab)* P(lab) > 1/2
(x/9) * (9-x/8) + (9-x/8) * (x/9) > 1/2
(9-x)x > 18
for this to be true,
x must be less than 6.
so this is sufficient to answer.
S-2
(9-x/9) * (8-x/8) >1/10
(9-x)(8-x) > 7.2
for this to be true, x must be less than 6.
so this is sufficient as well.
each option alone is sufficient.
Option D
Bunuel
16 Jun 2026, 16:27
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Peterparker2
Hi!
This would be the case if the problem asked for the probability not just for two different labradors, but for two different specific labradors. For example, if the labradors were A, B, C, D, E, F, G, H, I, and J, and the problem asked for the probability that you select H and J, then you would use this formula. Here’s why:
There are two scenarios in which you select H and J:
H first, then J
J first, then H
If we don’t multiply by 2!, then we only get the probability of one of those scenarios.
In this problem, we’re just looking for the probability of any two different labradors.
For more practice with Quant, check out ManhattanPrep’s Free GMAT QBank and Starter Kit.
Best,
SeedWasp
Joined: 08 May 2026
Last visit: 13 Jul 2026
Posts: 3
Given Kudos: 57
Location: India
Schools: ISB '26 Sloan '26 Sloan (MIT) Wharton '26
GPA: 9
17 Jun 2026, 22:15
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The no. of dogs is immaterial in this question.
At first glance, it feels like an algebraic trap. Your instinct is to start scratching out complex combinations using an unknown total number of dogs (N) and a total number of Labradors (L).
But you can completely skip the quadratic equations. Here is the elegant, 5-second conceptual trick that cracks this wide open—even if you have no idea how many dogs are in the shelter.
Whenever you grab two items from a mixed pile, there are only three possible outcomes in the entire universe. They must add up to exactly 1 (or 100%):
Now let's look at the statements using nothing but common sense.
Cracking Statement 1
Statement 1 tells us that the middle scenario—picking exactly one of each—is greater than 1/2.
Let's plug that in. If the middle scenario takes up more than half of the total probability pie, then the other two scenarios combined must share the remaining left-overs:
Probability of 2 Labs + Probability of 0 Labs < 1/2
Since you can't have a negative probability, P(2 Labs) is cleanly trapped. It has to be strictly less than 1/2.
Cracking Statement 2
Statement 2 tells us that picking zero Labradors is greater than 1/10 (10%).
Could the probability of picking two Labradors ever break past 1/2 (50%) here? Let's assume it could. If you have a massive 50% chance of pulling two Labradors, and a decent 10% chance of pulling zero Labradors, those two extremes alone consume 60% of your total probability.
In real-world math, when you mix two groups, you cannot have a huge probability at both extreme ends without the middle "mix" scenario expanding massively and spilling over the 100% total limit. The math breaks. Therefore, the probability of picking two Labradors is firmly locked down below half.
The Takeaway
Because both statements work perfectly on their own without knowing the total number of dogs, the correct answer is D (Each statement alone is sufficient).
At first glance, it feels like an algebraic trap. Your instinct is to start scratching out complex combinations using an unknown total number of dogs (N) and a total number of Labradors (L).
But you can completely skip the quadratic equations. Here is the elegant, 5-second conceptual trick that cracks this wide open—even if you have no idea how many dogs are in the shelter.
Whenever you grab two items from a mixed pile, there are only three possible outcomes in the entire universe. They must add up to exactly 1 (or 100%):
Now let's look at the statements using nothing but common sense.
Cracking Statement 1
Statement 1 tells us that the middle scenario—picking exactly one of each—is greater than 1/2.
Let's plug that in. If the middle scenario takes up more than half of the total probability pie, then the other two scenarios combined must share the remaining left-overs:
Probability of 2 Labs + Probability of 0 Labs < 1/2
Since you can't have a negative probability, P(2 Labs) is cleanly trapped. It has to be strictly less than 1/2.
- We get a guaranteed "Yes" to the question.
- Statement 1 is sufficient.
Cracking Statement 2
Statement 2 tells us that picking zero Labradors is greater than 1/10 (10%).
Could the probability of picking two Labradors ever break past 1/2 (50%) here? Let's assume it could. If you have a massive 50% chance of pulling two Labradors, and a decent 10% chance of pulling zero Labradors, those two extremes alone consume 60% of your total probability.
In real-world math, when you mix two groups, you cannot have a huge probability at both extreme ends without the middle "mix" scenario expanding massively and spilling over the 100% total limit. The math breaks. Therefore, the probability of picking two Labradors is firmly locked down below half.
- We get another guaranteed "Yes".
- Statement 2 is sufficient.
The Takeaway
Because both statements work perfectly on their own without knowing the total number of dogs, the correct answer is D (Each statement alone is sufficient).
Bunuel
