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QOTD # 17 Of the 40 specimens of bacteria in a dish, 3 speci [#permalink]
12 Sep 2016, 01:29
Question Stats:
58% (01:27) correct
41% (01:56) wrong based on 17 sessions
Of the 40 specimens of bacteria in a dish, 3 specimens have a certain trait. If 5 specimens are to be selected from the dish at random and without replacement, which of the following represents the probability that only 1 of the 5 specimens selected will have the trait? A) \(\frac{\frac{5}{1}}{\frac{40}{3}}=\) B) \(\frac{\frac{5}{1}}{\frac{40}{5}}=\) C) \(\frac{\frac{40}{3}}{\frac{40}{5}}=\) D) \(\frac{\frac{3}{1} \times \frac{37}{4}}{\frac{40}{3}}=\) E)\(\frac{\frac{3}{1} \times \frac{37}{4}}{\frac{40}{5}}=\) Practice Questions Question: 17 Page: 180
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Re: QOTD # 17 Of the 40 specimens of bacteria in a dish, 3 speci [#permalink]
12 Sep 2016, 01:46
ExplanationA question like this is much better to go step by step, dividing it into chunks. Now, we said that we have to select 1 specimens among five BUT only 3 over 40 have such traits. To select 5 among 40 is, the total possibilities, \(\frac{40}{5}\) in the denominator In the numerator, we have 1 specimens among 3 and then select 4 among 37 that do not have the traits. [m]\frac{\frac{3}{1} \times \frac{37}{4}}{\frac{40}{5}} The best answer is E
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Re: QOTD # 17 Of the 40 specimens of bacteria in a dish, 3 speci [#permalink]
13 Aug 2017, 09:17
I couldn't understand the solution properly.why we take 37C4??? Is that another way to solve this math ???



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Re: QOTD # 17 Of the 40 specimens of bacteria in a dish, 3 speci [#permalink]
13 Aug 2017, 13:13
The probability that only 1 of the 5 specimens selected from the 40 specimens will have the trait is equal to \(\frac{numbers of way to select 5 specimens, only 1 of which has the traits}{numbers of ways to select 5 specimens}\) The number of ways 5 specimens can be selected from the 40 specimens is \(\frac{40}{5}\).To select 5 specimens, only 1 of which has the trait, you have to select 1 of the 3 specimens that have the trait and select 4 of the 37 specimens that do not have the trait. The number of such selections is the product \(\frac{3}{1}\) * \(\frac{37}{4}\). So the probability that only 1 of the 5 specimens selected will have the trait is represented by \(\frac{3}{1}\) * \(\frac{37}{4}\) / \(\frac{40}{5}\) E is the best
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Re: QOTD # 17 Of the 40 specimens of bacteria in a dish, 3 speci [#permalink]
04 Jan 2019, 23:51
Carcass wrote: The probability that only 1 of the 5 specimens selected from the 40 specimens will have the trait is equal to
\(\frac{numbers of way to select 5 specimens, only 1 of which has the traits}{numbers of ways to select 5 specimens}\)
The number of ways 5 specimens can be selected from the 40 specimens is \(\frac{40}{5}\).To select 5 specimens, only 1 of which has the trait, you have to select 1 of the 3 specimens that have the trait and select 4 of the 37 specimens that do not have the trait. The number of such selections is the product \(\frac{3}{1}\) * \(\frac{37}{4}\).
So the probability that only 1 of the 5 specimens selected will have the trait is represented by \(\frac{3}{1}\) * \(\frac{37}{4}\) / \(\frac{40}{5}\)
E is the best I dont mean to correct a pro, but E is impossible. if you calculate E. that total is 3.46.... We are looking for a probability here, which has to be less than 1



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Re: QOTD # 17 Of the 40 specimens of bacteria in a dish, 3 speci [#permalink]
05 Jan 2019, 00:49
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Lets break it down muuuuch more simply.
First way is... the total number of combinations is going to be 40C5 which is 40!/(5!35!). this will be your denominator The total number of ways to pick one specimen with the desired train is 3C1, which is just 3, The total number of ways to pick four specimen without the desired train is 37C4 Now the question is asking what are the chances of picking a specimen with the certain trait, you just multiply 3*37C4, then divide that product by 40C5 3*37C4/(40C5)
which is (3*(37!/4!33!)/(40!/5!35!)



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Re: QOTD # 17 Of the 40 specimens of bacteria in a dish, 3 speci [#permalink]
05 Jan 2019, 12:00
Maybe the OA in the Official Quant Reasoning is wrong. However, as per the book , it is E
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Re: QOTD # 17 Of the 40 specimens of bacteria in a dish, 3 speci [#permalink]
28 Jan 2019, 09:14
Carcass wrote: Maybe the OA in the Official Quant Reasoning is wrong.
However, as per the book , it is E No, the OA is perfectly all right. There is typos in the answers here. For example for choice e, In instead of \(\frac{\frac{3}{1} * \frac{37}{4}}{\frac{40}{5}}=\), It should be \(\frac{3C1 * 37C4}{40C5}=\) It is the correct answer.



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Re: QOTD # 17 Of the 40 specimens of bacteria in a dish, 3 speci [#permalink]
28 Jan 2019, 12:06
No typo in the answer choices provided by the Offical Quantitative Reasoning Book Attachment:
shot35.jpg [ 17.18 KiB  Viewed 1767 times ]
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Re: QOTD # 17 Of the 40 specimens of bacteria in a dish, 3 speci [#permalink]
19 Mar 2019, 16:56
Carcass wrote: No typo in the answer choices provided by the Offical Quantitative Reasoning Book Attachment: shot35.jpg Does the format used in the official book IS the same as it has been used in the explanation ( on this forum, i.e. Numerator /Denominator form)??



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Re: QOTD # 17 Of the 40 specimens of bacteria in a dish, 3 speci [#permalink]
19 Mar 2019, 23:34
As AE pointed out, the answer should be in the format \(\frac{3C1 * 37C4}{40C5}=\) Regards
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Re: QOTD # 17 Of the 40 specimens of bacteria in a dish, 3 speci
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