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:

Re: Practice questions of Combinations and probability [#permalink]

Show Tags

05 Apr 2013, 04:51

Thank you for the collection. As I worked through the questions i noticed that some answers are not correct. For example answer 5 should be "B) 66" and not D. However, explaination is correct in my opinion. Cheers,

Re: Practice questions of Combinations and probability [#permalink]

Show Tags

16 Apr 2014, 03:06

Hello from the GMAT Club BumpBot!

Thanks to another GMAT 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.
_________________

Re: Practice questions of Combinations and probability [#permalink]

Show Tags

22 Sep 2014, 15:32

I have a doubt regarding Q4 in the doc: How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other?

The answer in the doc is 648 but it is derived on the basis - 9 X 9 X 8.

The question does not say the second and third digits are different from the first one. It says they are different from each other. Then should it not be 9 X 10 X 9? Am I mistaken?

I have a doubt regarding Q4 in the doc: How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other?

The answer in the doc is 648 but it is derived on the basis - 9 X 9 X 8.

The question does not say the second and third digits are different from the first one. It says they are different from each other. Then should it not be 9 X 10 X 9? Am I mistaken?

I see your point. Ambiguous wording. Ignore this question.
_________________

Re: Practice questions of Combinations and probability [#permalink]

Show Tags

28 Nov 2014, 17:38

Since 1 appears exactly three times, we can solve for the other four digits only. For every digit we can choose out of 8 digits only (without 1 and 0). Since we have 4 prime digits (2, 3, 5, 7) and 4 non-prime digits (4, 6, 8, 9), the probability of choosing a prime digit is ½. We need at least two prime digits: One minus (the probability of having no prime digits + having one prime digit): There are 4 options of one prime digit, each with a probability of (1/2)4. There is only one option of no prime digit with a probability of (1/2)4. So: [1- ((1/2)4+(1/2)4*4)] = 11/16. – Bernoulli’s principle

Hi Narenn This is in regards to your probability question bank 25 .

I dint quite get this - if there are three spots to be filled with 2 prime numbers - would 1/2(probablity of picking the first prime) * 3/7(probablity of picking the second one) would that not be enough - or are we assuming that there can be the two prime number which are same??

Re: Practice questions of Combinations and probability [#permalink]

Show Tags

09 Nov 2015, 12:04

Narenn wrote:

Hi All,

Below are Combination and Probability practice ques with complete solutions

Thanks,

Narenn

Hi Narenn, I've been reviewing this file, it's been great to enhance my weakest point on Quant (probability). Thanks a lot for that. Just one observation on questions 32:

The total number of diagonals is 171 (not 170) applying (21*(21-3)/2 - 18 = 171. It's not correct applying 20*(20-3)/2 since there will be an extra diagonal between points 1 and 20 that wouldn't exist in a polygon with 20 sides. Also 19*18/2=171

The same for question 33: (18*(18-3))/2 - (15+15+14) = 91. Or 13*14/2=91

Also please give me a little further explanation on question 46. I managed to solve it using the probability of not selecting a couple (1*(8/9)*(6/8)) = 2/3 multiplied by the total number of combination 10C3=120, (2/3)*120=80 . But I don't understand why 5C1*8C1=40 is the number of ways of selecting 3 individuals out of 5 couples.

Hope this helps to improve this great post. Regards

Re: Practice questions of Combinations and probability [#permalink]

Show Tags

30 Aug 2016, 20:57

Hi there,

Thanks for the post. This helps!

I think there is a problem in Q37.

37. How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?

a) 2520 b) 3150 c) 3360 d) 6000 e) 7500

Solution provided is:

The best answer is A. The first digit has 4 options (2,4,6,8 and not 0), the second has 5 options (1,3,5,7,9) the third has 3 options (3,5,7 and not 2), the fourth has 7 options (10-3 used before) and the fifth has 6 options (10-4 used before). The total is 4*5*3*7*6=2520.

I have problems with explanation in RED.

If on second and third position the number is repeated (3,5, or 7), then fourth position has 8 options and fifth position has 7 options.

So according to me, solution should be 4*2*3*7*6 + 4*3*3*8*7

Please correct me if i am wrong.

Pranav
_________________

What gets measured, gets managed

gmatclubot

Re: Practice questions of Combinations and probability
[#permalink]
30 Aug 2016, 20:57

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...