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06 Feb 2023, 07:43
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At a certain animal shelter, there are nine dogs available for adoption. If Susan chooses two dogs at random and p is the probability that Susan chooses two labradors, is p < 1/2 ?
(1) The probability of picking one labrador and one non-labrador is greater than 1/2.
(2) The probability that neither dog selected is a labrador is greater than 1/10.
(1) The probability of picking one labrador and one non-labrador is greater than 1/2.
(2) The probability that neither dog selected is a labrador is greater than 1/10.
10 Feb 2023, 16:29
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If there are 7 labradors, the probability of picking two labradors will be (7/9)(6/8) which is greater than 1/2, but if there are 6 (or fewer) labradors, the probability of picking two will be less than 1/2. So the question is really just asking if there are 6 or fewer labradors.
For Statement 1, the probability of picking one Lab and one non-Lab will be highest when we have lots of both. That's easy to see by imagining extremes: if we had 9 Labs and zero non-Labs, there'd be no chance we could pick one of each, whereas if we have roughly equal numbers of each, it will be fairly likely we get one of each. If we have 5 Labs and 4 non-Labs, or 4 Labs and 5 non-Labs, the probability of picking one of each is (5/9)(4/8) + (4/9)(5/8) which is greater than 1/2. If we have 6 of one, 3 of the other, the probability turns out to be (2)(6/9)(3/8) which is exactly 1/2. So Statement 1 tells us we have either 4 or 5 Labs, and is sufficient.
For Statement 2, if we had 7 Labs, the probability of picking 2 non-Labs would be (2/9)(1/8), which is less than 1/10. So we must have at least 3 non-Labs (and in fact we need to have at least 4, if you work out the probabilities), and thus at most 6 labradors (actually at most 5, if you do the math fully, but we don't need to), and Statement 2 is also sufficient, so the answer is D.
With these numbers, I'd never do this problem algebraically, but you can: for Statement 1, for example, if you imagine there are L Labradors, then there must be 9-L non-Labradors. The probability of picking one Lab and one non-Lab is then
(L/9)*[(9-L)/8] + [(9-L)/9]*(L/8) = (2)*(L)(9-L)/(9*8) = (L)(9-L)/36
We know this is greater than 1/2:
(L)(9-L)/36 > 1/2
(L)(9-L) > 18
L^2 -9L + 18 < 0
(L - 6)(L - 3) < 0
so one of these factors is negative, the other positive, and if that's true, L-6 must be negative (because it's smaller than L-3), so L-6 < 0 and L-3 > 0, and 3 < L < 6. One reason though to avoid the algebra in situations like this is that the quadratic we arrive at here is often not factorable without the quadratic formula (it's lucky coincidence, or elegant question design, that it is in this case), so you often end up just imagining numerical situations anyway.
For Statement 1, the probability of picking one Lab and one non-Lab will be highest when we have lots of both. That's easy to see by imagining extremes: if we had 9 Labs and zero non-Labs, there'd be no chance we could pick one of each, whereas if we have roughly equal numbers of each, it will be fairly likely we get one of each. If we have 5 Labs and 4 non-Labs, or 4 Labs and 5 non-Labs, the probability of picking one of each is (5/9)(4/8) + (4/9)(5/8) which is greater than 1/2. If we have 6 of one, 3 of the other, the probability turns out to be (2)(6/9)(3/8) which is exactly 1/2. So Statement 1 tells us we have either 4 or 5 Labs, and is sufficient.
For Statement 2, if we had 7 Labs, the probability of picking 2 non-Labs would be (2/9)(1/8), which is less than 1/10. So we must have at least 3 non-Labs (and in fact we need to have at least 4, if you work out the probabilities), and thus at most 6 labradors (actually at most 5, if you do the math fully, but we don't need to), and Statement 2 is also sufficient, so the answer is D.
With these numbers, I'd never do this problem algebraically, but you can: for Statement 1, for example, if you imagine there are L Labradors, then there must be 9-L non-Labradors. The probability of picking one Lab and one non-Lab is then
(L/9)*[(9-L)/8] + [(9-L)/9]*(L/8) = (2)*(L)(9-L)/(9*8) = (L)(9-L)/36
We know this is greater than 1/2:
(L)(9-L)/36 > 1/2
(L)(9-L) > 18
L^2 -9L + 18 < 0
(L - 6)(L - 3) < 0
so one of these factors is negative, the other positive, and if that's true, L-6 must be negative (because it's smaller than L-3), so L-6 < 0 and L-3 > 0, and 3 < L < 6. One reason though to avoid the algebra in situations like this is that the quadratic we arrive at here is often not factorable without the quadratic formula (it's lucky coincidence, or elegant question design, that it is in this case), so you often end up just imagining numerical situations anyway.
General Discussion
Bunuel
(1) The probability of picking one labrador and one non-labrador is greater than 1/2.
Even if have 8 Labrodors and 1 other we cannot get p> \(\frac{1}{2}\)
\(\frac{8}{9}*\frac{1}{8}=\frac{1}{9} < \frac{1}{2} \)
It seems it is not possible to get this statement as true using any combination of Labradors and others
p seems always < \(\frac{1}{2}\) using any # of Labradors
(2) The probability that neither dog selected is a labrador is greater than 1/10.
If we have 4 Labradors and 5 other then:
\(\frac{5}{9}*\frac{4}{8}=\frac{5}{18} > \frac{1}{10}\)
We can answer Yes to the stem
If we have 5 Labradors and 4 other then:
\(\frac{4}{9}*\frac{3}{8}=\frac{1}{6} > \frac{1}{10}\)
We can answer NO to the stem
INSUFF.
Bunuel can you please check the question. Thank you.
