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30 Apr 2026, 19:00
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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.
The OA will be automatically revealed on Monday 4th of May 2026 07:00:13 PM Pacific Time Zone
(1) Fewer than half of the animals were cats.
(2) The probability that one dog and one cat are selected is 16/33.
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The OA will be automatically revealed on Monday 4th of May 2026 07:00:13 PM Pacific Time Zone
30 Apr 2026, 22:20
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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
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
01 May 2026, 05:16
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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)
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)
