At an animal shelter, 12 animals were available for adoption.

Question

At an animal shelter, 12 animals were available for adoption. Some were dogs, and the rest were cats. If two animals are selected at random, without replacement, is the probability that both selected animals are dogs greater than 1/3?

(1) Fewer than half of the animals were cats.
(2) The probability that one dog and one cat are selected is 16/33.

M47-135

Bunuel · Accepted Answer

GMAT Club Official Solution:

At an animal shelter, 12 animals were available for adoption. Some were dogs, and the rest were cats. If two animals are selected at random, without replacement, is the probability that both selected animals are dogs greater than 1/3?

Let d be the number of dogs. Then the question asks:

Is d/12 * (d - 1)/11 > 1/3?
Is d(d - 1) > 44?

By trial and error we can get that this is equivalent to asking whether d > 7.

(1) Fewer than half of the animals were cats.

This means the number of cats was less than 6. Thus, d > 6.

Not sufficient.

(2) The probability that one dog and one cat are selected is 16/33.

The probability of selecting one dog and one cat is

d/12 * (12 - d)/11 * 2! = 16/33
d(12 - d) = 32

This gives d = 4 or d = 8. One gives a NO answer, the other gives a YES answer.

Not sufficient.

(1) + (2) From (1), d > 6, so from (2), d can only be 8. Therefore, the answer is yes. Sufficient.

Answer: C.

rajdeep212003 · Answer

Let dogs = d
So cats = 12 − d

We need:

P(2 dogs) = d(d−1) / (12×11) > 1/3

---

Statement (1):

Fewer than half are cats
→ cats < 6 → dogs > 6

So possible values: d = 7,8,9,10,11,12

Try:

d = 7 → 42/132 < 1/3
d = 8 → 56/132 > 1/3

Not consistent → not sufficient

---

Statement (2):

P(1 dog, 1 cat) = 16/33

So:

d(12−d) / 66 = 16/33

→ d(12−d) = 32
→ d2 − 12d + 32 = 0
→ (d−4)(d−8) = 0

So d = 4 or 8

Check:

d = 4 → 12/132 < 1/3
d = 8 → 56/132 > 1/3

Again not fixed → not sufficient

---

Both together:

From (1): d > 6
From (2): d = 4 or 8

Common value → d = 8

Now:

56/132 > 1/3 → works

---

Answer: C

---

Hope this helps
Feel free to tag me if needed

– Rajdeep

SS990 · Answer

P(d) = Probability of choosing a dog
Probability of choosing 2 dogs is 12C2= 12 x 11/ 2 x 1 = 66

to find if P(d)>= 1/3 , multiplying 22 up and down to get 66 in denominator,  then P(d)>= 22/66 
so dC2>=22 i.e d>=8 cos if d=7, then 7C2 is 21which wont satisfy the question criteria

1) fewer than half of animals (12/2 = 6) is cats, so cats can be 5,4,3,2,1 and dogs can be 7,8,9,10,11 resp...so not sufficient to prove if d>=8 or not

2) P (1d & 1c) = 16/33 0r 32/66 = 8 x 4/66 , but we dont know if 8 is cats or dogs.. so insufficient

both together we know cat is less than half and 8 x 4 is combinator of cxd so we can answer the Question criteria.

Hence (c)

BongBideshini · Answer

Let: Dogs = d Cats = c Total: d + c = 12 Need to know whether the probability of selecting 2 dogs is greater than 1/3. Probability of 2 dogs: dC2 / 12C2 Since: 12C2 = 66 Need: dC2 / 66 > 1/3 dC2 > 22 d(d - 1)/2 > 22 d(d - 1) > 44 Is d >= 8? Statement (1) c< 6 Dogs > 6 So d could be: 7, 8, 9, 10, 11, 12 Need d ≥ 8 Could be 7 => No Could be 8 => Yes Not sufficient Statement (2) (d × c) / 66 = 16/33 d × (12-d) = 32 d = 8 or 4 If d = 8, condition satisfied-> yes d=4, condition does not satisfied-> no Not Sufficient Together (1) + (2) c<6 , means d>6 So, d = 8. Ans (C)

Adit_ · Answer

Statement 1:
When we take 7 dogs and 5 cats:
We have probability of 7/22 for both being dogs <1/3
Any greater no.of dogs we will clearly see this probability being >1/3.
INSUFFICIENT.

Statement 2:
First lets extrapolate the fraction:
16/33 = 64/132. [Since we know we will always have 12*11 form in the denominator.]
Now choosing one dog and one cat can be done in two ways for each pair:
8+4 or 4+8.
Each cases when added up gives us a total probability of 64/132 [32/132+32/132]
So now we know number of dogs is either 4 or 8.
For 4 it is <1/3.
For 8 it is >1/3.
INSUFFICIENT.

Combining both:
Now we have exactly what we need.
First statement says no.of cats is less than half.
When we combine this with statement 2 we see that no.of dogs must be 8 only.
Probability >1/3 here.
Answer is YES.
SUFFICIENT.

Answer: Option C

_________________________

Bunuel since the official answer is already uploaded could you release the results so we know the actual difficulty of the problem and the success rate?

Bunuel · Answer

Here it is: https://gmatclub.com/forum/download/file.php?id=88409 GMAT-Club-Forum-3vdoalz6.png

At an animal shelter, 12 animals were available for adoption.

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