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 350,000 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: Question about probability? [#permalink]
21 Jan 2014, 11:49

a03b wrote:

From a bag of 6 candies: 4 sweet, 2 sour Jack picks 2 candies simultaneously. What is the probability that first candy is sweet, second is sour?

My book says 4/6 x 2/5 = 4/15

The part I don't understand is why picking simultaneously and picking in a sequence are considered the same. There should be a difference.

Thanks for any replies.

Perhaps you should be asking - if you're picking them simultaneously, why is there a first and second? Sorry, that wasn't helpful.

Imagine if the first candy you pick is sour, and then the second candy you pick is sweet. Then what's the probability? It should be 2/6 x 4/5 = 4/15.

As you can see, the order doesn't actually matter. So whether they're picked simultaneously or picked in a sequence, the answer is the same.

The reason why they specified "simultaneously" is probably because they don't want you throwing the candy back into the bag after you pick it. (probability with replacement is different than probability without replacement) _________________

Re: Question about probability? [#permalink]
21 Jan 2014, 12:03

Hmm... So what about this question? Is it 1/9 or 2/9. I'm so confused.

There are 3 red, 2 white, and 5 blue marbles. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marbles?

Re: Question about probability? [#permalink]
21 Jan 2014, 13:14

a03b wrote:

Hmm... So what about this question? Is it 1/9 or 2/9. I'm so confused.

There are 3 red, 2 white, and 5 blue marbles. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marbles?

So what's the probability that the first marble we pick is white? There are 2 white marbles among 10 total marbles, so the probability is 2/10. After we pick out this white marble, what's the probability that the next marble we pick is blue? There are 5 blue marbles from 9 total marbles, so the probability is 5/9.

So, 2/10 x 5/9 = 1/9.

If we reverse the order of picking, and pick the blue marble first, then it becomes 5/10 x 2/9 = 1/9.

Re: Question about probability? [#permalink]
21 Jan 2014, 13:17

farful wrote:

a03b wrote:

Hmm... So what about this question? Is it 1/9 or 2/9. I'm so confused.

There are 3 red, 2 white, and 5 blue marbles. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marbles?

So what's the probability that the first marble we pick is white? There are 2 white marbles among 10 total marbles, so the probability is 2/10. After we pick out this white marble, what's the probability that the next marble we pick is blue? There are 5 blue marbles from 9 total marbles, so the probability is 5/9.

So, 2/10 x 5/9 = 1/9.

If we reverse the order of picking, and pick the blue marble first, then it becomes 5/10 x 2/9 = 1/9.

Re: Question about probability? [#permalink]
21 Jan 2014, 13:22

a03b wrote:

farful wrote:

a03b wrote:

Hmm... So what about this question? Is it 1/9 or 2/9. I'm so confused.

There are 3 red, 2 white, and 5 blue marbles. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marbles?

So what's the probability that the first marble we pick is white? There are 2 white marbles among 10 total marbles, so the probability is 2/10. After we pick out this white marble, what's the probability that the next marble we pick is blue? There are 5 blue marbles from 9 total marbles, so the probability is 5/9.

So, 2/10 x 5/9 = 1/9.

If we reverse the order of picking, and pick the blue marble first, then it becomes 5/10 x 2/9 = 1/9.

So same answer, either way.

When we add, the result is 2/9 though. Right?

Posted from my mobile device

Sorry if I've confused you. The answer is 1/9. I only did it both ways to show that order of sequence does not matter. There's no reason to add anything. _________________

Re: Question about probability? [#permalink]
22 Jan 2014, 11:59

But there are C(10,2)=45 ways to choose 2 balls out of 10. There are 2x5=10 desired out comes. 10/45 = 2/9 I'm so confused now. If the answer to this question is 1/9, I have some really serious problems.

Re: Question about probability? [#permalink]
22 Jan 2014, 12:06

a03b wrote:

But there are C(10,2)=45 ways to choose 2 balls out of 10. There are 2x5=10 desired out comes. 10/45 = 2/9 I'm so confused now. If the answer to this question is 1/9, I have some really serious problems.

Ugh, I'm terribly sorry. You are correct, the answer should be 2/9.

We should add the 1/9 and 1/9 I got from my previous answers because there are two ways to get 1 white and 1 blue... choosing the white first then blue, or choosing the blue first then white.

I feel I may have made this whole thing more confusing for you. My apologies. _________________

Re: Question about probability? [#permalink]
22 Jan 2014, 12:18

No problem farful, thank you for your replies. Since I'm not a native speaker, now all simultaneous picking, picking in order, etc... are much more clear. Cheers.

Re: Question about probability? [#permalink]
22 Jan 2014, 12:19

a03b wrote:

From a bag of 6 candies: 4 sweet, 2 sour Jack picks 2 candies simultaneously. What is the probability that first candy is sweet, second is sour?

My book says 4/6 x 2/5 = 4/15

The part I don't understand is why picking simultaneously and picking in a sequence are considered the same. There should be a difference.

Thanks for any replies.

Let me reanswer this.

The reason why they said 'simultaneously' is because they wanted you to pick one candy, then pick a second candy without putting the first candy back into the bag.

So they really want the probability of picking a sweet candy first, followed by the probability of picking a sour candy next (from the remaining 5).

If the question was: From the bag, pick two candies, what is the probability one is sweet and one is sour? Then, the answer would be 8/15. You can get this answer either 1) via (4x2)/(2choose6) or 2) figuring out the probability of picking sweet then sour and adding that to the probability of picking sour then sweet, since there are two ways to get one sweet and one sour.

Great to know you are joining Kellogg. A lot was being talked about your last minute interview on Pagalguy (all good though). It was kinda surprise that you got the...

This is a long overdue post! A lot of Indian applicants, having scheduled interviews in March, reached out to me asking about my interview experience with Kellogg. I had a...

A critical phase of the MBA application, concurrent to researching your target schools, is “researching yourself” and building your profile. What are your unique traits? Where do you want to be...