The number of cats in Libbys house is twice the number of dogs and on

Question

The number of cats in Libby’s house is twice the number of dogs and one third the number of fish. If cats, dogs and fish are the only pets in Libby’s house, what is the probability of randomly selecting two cats?

(1) The number of cats in Libby’s house is 12.
(2) The total number of pets in Libby’s house is 54.

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dhart · Answer

looks like option d , both the statements looks sifficient

odionam · Answer

C=2D=F/3 we have to find probability of 2 cats selected randomly St1= c=12 we can find D=6 and F=36 so we can find probability and st2 total is given we have ratio of C: D: F= 2:1:6 so we can find individual from this and calculate probability so answer is D each alone is sufficient

eteum · Answer

If the stem questions already gives you the ratios of all the animals. Can't you already solve the problem without even looking at the 2 additional statements?

probability of choosing a cat is:

c / (c + c/2 + 3c) = c / 4.5c which is 2/9

egmat · Answer

Hi eteum,

Your c / 4.5c =  2/9  is completely correct, and it's worth seeing  why  it works, because that is the exact spot where this question turns.

Write every count as a multiple of one unknown: cats =  2k , dogs =  k , fish =  6k , total =  9k . For  one  cat:

P = 2k / 9k =  2/9

The  k cancels  - top and bottom both scale with the size of the house, so the ratio alone genuinely is enough here. Your reasoning is sound.

Two cats is a different animal (no pun intended)

Once you have picked the first cat, that cat is gone. So:

P(two cats) = (2k / 9k) x (2k - 1) / (9k - 1) = (2/9) x (2k - 1) / (9k - 1)

Look at that second fraction. The  -1  does not scale with k, so  k refuses to cancel . Removing one pet changes the mix of what is left - and  how much  it changes depends on how many pets there were to begin with. That is why the ratio stops being enough.

That is also all the combinations formula C(cats, 2) / C(total, 2) is doing: counting unordered pairs of cats over unordered pairs of pets. Same thing, with the  -1  hidden inside.

Two cases that prove it

Both obey the stem's ratio:

- k =  1 : cats  2 , total  9  - so (2/9)(1/8) =  1/36 , about  0.028 
- k =  6 : cats  12 , total  54  - so (2/9)(11/53) =  22/477 , about  0.046

Same ratio,  two different probabilities , so the stem alone is  not sufficient . You need k.

And that is exactly what each statement hands you:

- (1) cats =  12 , so k =  6 .  Sufficient. 
- (2) total =  54  = 9k, so k =  6 .  Sufficient.

Each alone pins down k,  therefore  the answer is  D .

Rule to carry:  a ratio settles a  single -draw probability, but with  two draws without replacement  the -1 breaks the cancellation, and that needs actual counts.

Answer: D

The number of cats in Libbys house is twice the number of dogs and on

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GMAT Data Sufficiency (DS) Questions

The number of cats in Libbys house is twice the number of dogs and on

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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