Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

GMAT Diagnostic Test Question 37 Field: probability Difficulty: 700

A box has 6 red hats and 5 green hats. What is the probability of drawing at least one green hat in two consecutive drawings if the hat is not replaced?

A. \(\frac{10}{11}\) B. \(\frac{8}{11}\) C. \(\frac{7}{12}\) D. \(\frac{5}{13}\) E. \(\frac{2}{7}\)
_________________

I think this is the best approach. Especially when there are multiple situations that satisfy the question. In this question, it was simple, there are only 4 possble results RR, RG, GR, GG, and the only one that doesn't satisfy the solution is RR. So rather than finding the probability of RG, GR, and GG that DOES satisfy the "at least 1 green hat". If we find the situations that DO satisfy, meaning all 3 situations, then we have to add together their total probability. If we find the probability of the 1 situation we DO NOT want, all we have to do is subtract that probability from 1 and it's a much simpler problem. I would say much simpler than a 700 level question.

hrish88 wrote:

For me easiest way is 1-(RR)=1-6/11*5/10=8/11

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

The probability of two green balls in succession (going by 1 - Probability of 2 red balls) is 8/11 and this makes sense. However, cant we get the calculation of p(1st green ball) = 5/11 and probability of drawing 2nd green ball (p 2nd green ball) = 4/10 so 5/11 * 4/10 = 2/11. What is wrong with this approach?

This is wrong.

See you have three cases. RG, GR , GG i.e. one of them is green or both of them is green.

P(RG) = P(GR) = 6/11 * 5/10 = 3/11 -->You have missed this case

P(GG) = 5/11 * 4/10 = 2/11 -->You have considered only this case

total probability = P(RG) + P(GR) + P(GG) = 3/11 + 3/11 + 2/11 = 8/11
_________________

There are 11 hats in a box, out of which 6 are red hats and 5 are green hats. If two hats are to be selected at random without replacement, what is the probability that at least one green hat will be selected?

A. \(\frac{10}{11}\) B. \(\frac{8}{11}\) C. \(\frac{7}{12}\) D. \(\frac{5}{13}\) E. \(\frac{2}{7}\)

It's easier to find the probability of the opposite event and subtract it from 1. The opposite event would be if we select zero green hats, or which is the same if we select 2 red hats, so: \(P(G\geq{1})=1-P(RR)=1-\frac{6}{11}*\frac{5}{10}=\frac{8}{11}\).

The probability of two green balls in succession (going by 1 - Probability of 2 red balls) is 8/11 and this makes sense. However, cant we get the calculation of p(1st green ball) = 5/11 and probability of drawing 2nd green ball (p 2nd green ball) = 4/10 so 5/11 * 4/10 = 2/11. What is wrong with this approach?

The way i did was, 3 possibilities, (A) both drawn are green:2/11 (B) 1st green 2nd red :3/11 (C) 1st red 2nd green :3/11. Hence total probability of atleast 1 green=P(A)+P(B)+P(C)=8/11.

I think perhaps one of the answer choices should be 3/11, a trick to catch someone not paying 100% attention is one I see frequently on other questions. Thanks for the test, really helpful.

Easiest way to do this type of question is to analize the worst case scenario

If you want the probability of drawing at least one green hat in two consecutive drawings, first count the possible outcomes drawing red hats

1 red hat, 1 rh, 1 rh, 1 rh, 1 rh, 1 rh.

Ok, so now we have exausted all possible outcomes without drawing any green hat. Subsequent to this count (1 red hat, 1 rh, 1 rh, 1 rh, 1 rh, 1 rh), start drawing green hats until you have the desired result, which is at least 2 concecutive green hats. This final count adds up to 8 drawings. Finally, divide 8 drawings by the possible number of hats in total. So, the probability of drawing at least one green hat in two consecutive drawings is 8/11.

Okay, I get the inverse probability idea, but let's say you do this,

P(GG) + P(GR) + P(RG) = correct answer I know that the correct answer is 8/11, which you can get to like this: P(GG) = 5/11 * 4/10 = 20/110 P(GR) = 5/11 * 6/10 = 30/110 P(RG) = 6/11 * 5/10 = 30/110 so correct answer = 80/110 = 8/11

However, I'm weak on the concept of P(GG) and could use some strengthening there. Why don't you multiply 20/110 by 2 to reflect that GG could be ordered either way?

Okay, I get the inverse probability idea, but let's say you do this,

P(GG) + P(GR) + P(RG) = correct answer I know that the correct answer is 8/11, which you can get to like this: P(GG) = 5/11 * 4/10 = 20/110 P(GR) = 5/11 * 6/10 = 30/110 P(RG) = 6/11 * 5/10 = 30/110 so correct answer = 80/110 = 8/11

However, I'm weak on the concept of P(GG) and could use some strengthening there. Why don't you multiply 20/110 by 2 to reflect that GG could be ordered either way?

GR and RG are different cases: G first than R and R first than G.