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Among all the students at a certain high school, the probabi
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02 Nov 2019, 02:49

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Question Stats:

Among all the students at a certain high school, the probability of picking a left-handed student is \(\frac{1}{4}\), and the probability of picking a student who is learning Spanish is \(\frac{2}{3}\) Which of the following could be the probability of picking a student who is either left-handed or learning Spanish or both?

Indicate all such numbers.

A. \(\frac{1}{2}\)

B. \(\frac{2}{3}\)

C. \(\frac{3}{4}\)

D. \(\frac{5}{6}\)

E. \(\frac{7}{8}\)

Indicate all such numbers.

A. \(\frac{1}{2}\)

B. \(\frac{2}{3}\)

C. \(\frac{3}{4}\)

D. \(\frac{5}{6}\)

E. \(\frac{7}{8}\)

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Official Answer

B,C,D,E

Re: Among all the students at a certain high school, the probabi
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23 Jan 2020, 04:49

7

huda wrote:

Among all the students at a certain high school, the probability of picking a left-handed student is \(\frac{1}{4}\), and the probability of picking a student who is learning Spanish is \(\frac{2}{3}\) Which of the following could be the probability of picking a student who is either left-handed or learning Spanish or both?

Indicate all such numbers.

A. \(\frac{1}{2}\)

B. \(\frac{2}{3}\)

C. \(\frac{3}{4}\)

D. \(\frac{5}{6}\)

E. \(\frac{7}{8}\)

Indicate all such numbers.

A. \(\frac{1}{2}\)

B. \(\frac{2}{3}\)

C. \(\frac{3}{4}\)

D. \(\frac{5}{6}\)

E. \(\frac{7}{8}\)

WhatsApp Image 2020-01-23 at 6.19.33 PM.jpeg [ 62.84 KiB | Viewed 9469 times ]

Re: Among all the students at a certain high school, the probabi
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09 Jul 2020, 02:05

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please explain.

Re: Among all the students at a certain high school, the probabi
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09 Jul 2020, 03:52

1

Expert Reply

The explanation above is quite net.

What is still not clear ??

Regards

What is still not clear ??

Regards

Re: Among all the students at a certain high school, the probabi
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17 Aug 2020, 14:59

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1

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huda wrote:

Among all the students at a certain high school, the probability of picking a left-handed student is \(\frac{1}{4}\), and the probability of picking a student who is learning Spanish is \(\frac{2}{3}\) Which of the following could be the probability of picking a student who is either left-handed or learning Spanish or both?

Indicate all such numbers.

A. \(\frac{1}{2}\)

B. \(\frac{2}{3}\)

C. \(\frac{3}{4}\)

D. \(\frac{5}{6}\)

E. \(\frac{7}{8}\)

Indicate all such numbers.

A. \(\frac{1}{2}\)

B. \(\frac{2}{3}\)

C. \(\frac{3}{4}\)

D. \(\frac{5}{6}\)

E. \(\frac{7}{8}\)

For this type of question, it'd be better to visualize it with a venn diagram as RSQUANT has demonstrated above. It would also help to pick a number of total students at the school, namely one that is a multiple of 3 and 4. This question tests your understanding of overlapping groups/probability.

___________________________

If you already understand the concepts, skip to below the italicized for the answer.

For starters, what is the question asking? It's asking for all of the students that are in spanish, OR are left handed, \(OR\) are in both. The OR represents either Spanish student is picked, OR a left handed student is picked, OR a Spanish student AND a left-handed student is picked. The distinction between AND and OR is a very important concept to grasp for word problems such as these.

For this question, using the visual of a venn diagram enclosed in a box, we are focused on everything but the area outside the venn diagram (the neither area). This is RSQUANT's first venn diagram.

___________________________

Continuing, we have \(\frac{2}{3}\) of the students can be taking spanish and \(\frac{1}{4}\) can be lefted handed. How do we maximize/minimize the group of both?

I'll be picking 60 as the total number of students. Well, \(\frac{2}{3}\) of 60 is 40 students and \(\frac{1}{4}\) of 60 is 15 students.

So what if ALL of the students that are left handed take spanish?

Then that would mean 15 students are left handed AND take spanish. Notice that this is the both group.

Therefore, in this scenario, the probability of picking a left handed student OR a Spanish student OR both is 40 out of 60 students, since there are 40 students taking spanish, and the 15 left-handed students are subsumed within the group of spanish students (RSQUANTS's third venn diagram).

So we have a minimum of \(\frac{40}{60}\) \(=\) \(\frac{2}{3}\)

______________________________

In the next scenario, let's assume that NONE of the left handed students are taking spanish.

That means there is NO overlap between left handed students and spanish students, and in fact, we have some students that are NEITHER left handed or taking spanish.

This is equivalent to both circles in the venn diagram being seperate, as the second venn diagram of RSQUANT shows. Mathematically, this can be seen this way: 40+15 = 55, but we have 60 students total, so those extra 5 students must be right handed and not taking spanish.

Now, how many students take spanish, OR are left-handed, OR take both? We know that's 55 out of 60 students, which is \(\frac{11}{12}\).

\(\frac{11}{12}\) is the maximum.

______________________________

So now we have our range: \(\frac{2}{3}\) <= P[S OR L OR (S AND L)] <= \(\frac{11}{12}\).

B,C,D, and E all fall within this range, so they are the answers.

Re: Among all the students at a certain high school, the probabi
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01 Sep 2020, 10:50

In this question you have to think in the best and worst situation. The best situation is when you have more probabilities and you will have more probabilities if is independent, I mean when there is no a both left-hand and Spanish learning student. In this case you will have a probability of A+B-A*B = 1/4 + 2/3 - 0 = 11/12. The other situation is when you have the lowest probabilities and this is when there are the highest number of both left-hand and Spanish students. In this case the total probabilities will be the the highest number of independent probabilities in this case 2/3. So the probabilities of picking a left-hand or Spanish or both will range from 2/3 to 11/12. B,C,D,E are correct answers.

Re: Among all the students at a certain high school, the probabi
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03 Feb 2024, 09:17

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

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