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:

Q: If the probability of a certain event is p, what is the probability of it occurring k times in n-time sequence? (Or in English, what is the probability of getting 3 heads while tossing a coin 8 times?) Solution: All events are independent. So, we can say that:

\(P' = p^k*(1-p)^{n-k}\) (1)

But it isn't the right answer. It would be right if we specified exactly each position for events in the sequence. So, we need to take into account that there are more than one outcomes. Let's consider our example with a coin where "H" stands for Heads and "T" stands for Tails: HHHTTTTT and HHTTTTTH are different mutually exclusive outcomes but they both have 3 heads and 5 tails. Therefore, we need to include all combinations of heads and tails. In our general question, probability of occurring event k times in n-time sequence could be expressed as:

\(P = C^n_k*p^k*(1-p)^{n-k}\) (2)

In the example with a coin, right answer is \(P = C^8_3*0.5^3*0.5^5 =C^8_3*0.5^8\)

I didn't get how we have found 0.5. The power 3 - is it mean the number of needed H and the power 5 represent T? So, what is 0.5?

Example #1 Q.:If the probability of raining on any given day in Atlanta is 40 percent, what is the probability of raining on exactly 2 days in a 7-day period? Solution: We are not interested in the exact sequence of event and thus apply formula #2: \(P = C^7_2*0.4^2*0.6^5\)

I want to specify, the powers 2 and 5 are the numbers of days?? I assume that 0,4 it's the likelihood of the raining and 0.6 of the sunny weather. Right?

I didn't get how we have found 0.5. The power 3 - is it mean the number of needed H and the power 5 represent T? So, what is 0.5?

0.5 is probability of getting head (or tail).

fruit wrote:

I want to specify, the powers 2 and 5 are the numbers of days?? I assume that 0,4 it's the likelihood of the raining and 0.6 of the sunny weather. Right?

Example #1 Q.:If the probability of raining on any given day in Atlanta is 40 percent, what is the probability of raining on exactly 2 days in a 7-day period? Solution: We are not interested in the exact sequence of event and thus apply formula #2: \(P = C^7_2*0.4^2*0.6^5\)

For some reason this equation form confuses me when I see it written:

\(P = C^7_2*0.4^2*0.6^5\)

Because the events on the 7 days are independent (mathematically)

I think of it in the following manner \(P = C^7_2*p(rain)*p(rain)*p(norain)*p(norain)*p(norain)*p(norain)*p(norain)\)

Can anyone think of a question where this will give me headaches in the future?

I find it much easier to remember that after \(C^7_2\) I just use the probability of the events that I want to occur.

Example #2 Q: Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other? Solution:

1) combinatorial approach: \(C^5_3\) - we choose 3 couples out of 5 couples. \(C^2_1\) - we chose one person out of a couple. \((C^2_1)^3\) - we have 3 couple and we choose one person out of each couple. \(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

Reading the OG12 math section right now -- does anybody know why it bothers to mathematically define mutually exclusive/independent events? I'm a little confused by its purpose. (Page 119 for those who are curious).

They take the intersection of A and B and divide it by the # of times an event A can occur. Which, for mutually exclusive sets, spits back out the same probability. But they aren't using the whole number of possible outcomes to arrive at this possibility, which is confusing me a bit.

In the bigger picture, though, why are they even including this in the first place? Are there any non-independent probabilities that we have to solve through?
_________________

Walker: Do the Gmac always mentions the replacement of the balls? If they have not mentioned what should be taken by default? No replacement or replacement?
_________________

Walker: Do the Gmac always mentions the replacement of the balls? If they have not mentioned what should be taken by default? No replacement or replacement?

I've never seen any official GMAC question with ambiguity. So, I'm pretty sure that GMAC will say directly replacement/no replacement if it's important.
_________________

Example #1 Q: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly chosen to form a committee, what is the probability that the committee includes both Bob and Rachel? Solution:

1) combinatorial approach: The total number of possible committees is N=C^8_2. The number of possible committee that includes both Bob and Rachel is n=1. P = \frac{n}{N} = \frac{1}{C^8_2} = \frac{1}{28}

2) reversal combinatorial approach: Instead of counting probability of occurrence of certain event, sometimes it is better to calculate the probability of the opposite and then use formula p = 1 - q. The total number of possible committees is N=C^8_2. The number of possible committee that does not includes both Bob and Rachel is: m = C^6_2 + 2*C^6_1 where, C^6_2 - the number of committees formed from 6 other people. 2*C^6_1 - the number of committees formed from Rob or Rachel and one out of 6 other people.Why is it multiplied by 2? Can someone pls explain it to me? P = 1- \frac{m}{N} = 1 - \frac{C^6_2 + 2*C^6_1}{C^8_2} P = 1 - \frac{15+2*6}{28} = 1 - \frac{27}{28} = \frac{1}{28}
_________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done." -Nelson Mandela

\(C^6_1\) - the number of committees formed from Bob and one person out of 6 people. \(C^6_1\) - the number of committees formed from Rachel and one person out of 6 people.

So, \(C^6_1 + C^6_1 = 2*C^6_1\) - the number of committees formed from Rob or Rachel and one person out of 6 other people.

Thanks Walker! Great post
_________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done." -Nelson Mandela

Iam having difficulty solving this. I was thinking on lines of conditional probability.

The probability of a day having 90% humidity on any day of may is x. The probability that lightening will occur on a humid day of may is y. The probability that lightening may occur on any day of may is z. What is the probability that lightening will occur on a day of may?

Need some lucid explanation on such problems. I've been having difficulties with probability problems of lately.

I'm having a little trouble with the solution to a sample problem in MGMAT Word Translations (3rd Ed) p. 99-100.

Q: "A medical researcher must choose one of 14 patients to receive an experimental medicine called Progaine. The researcher must then choose one of the remaining 13 patients to receiver another medicine, called Ropecia. Finally, the researcher administers a placebo to one of the remaining 12 patients. All choices are equally random. If Donald is one of the 14 patients, what is the probability that Donald receivers either Prograine or Ropecia?"

I thought this was a simple "without replacement" question.

Probability of Donald getting Progaine: 1/14 Probability of Donald getting Ropecia after one patient is removed from the pool: 1/13

(1/14)+(1/13) = 27/182

The book says the answer is

Probability of Donald getting Prograine: 1/14 Probability of Donald getting Ropecia: 1/14

(1/14)+(1/14) = 1/7

Anyone see why this is not a "without replacement" problem? I'm stumped.

Donald Getting prograine = 1/14 (1 out of 14 patients should be chosen for prograine) Donald Not getting prograine = 13/14 Donald Getting Ropecia = 1/13 (1 out of 13 patients should be chosen for Ropecia)

How can Donald get either P or R Donald will get Prograine and get done OR Donald will not get Prograine and get Ropecia

1/14+ 13/14*1/13 = 1/14+1/14 = 1/7

I believe it is a without replacement problem.

Thanks! I see the step I was missing now. For Donald to get Ropecia, not only does one one person have to be taken away from the pool (making his chances now 1/13), but I also have to factor in the 13/14 because it was not Donald who was the patient who received prograine. Thank you very much!

“Sure I am this day we are masters of our fate, that the task which has been set before us is not above our strength; that its pangs and toils are not beyond our endurance. As long as we have faith in our own cause and an unconquerable will to win, victory will not be denied us.” - Winston Churchill

Bunuel, really a fan of ur math book..
_________________

“Sure I am this day we are masters of our fate, that the task which has been set before us is not above our strength; that its pangs and toils are not beyond our endurance. As long as we have faith in our own cause and an unconquerable will to win, victory will not be denied us.” - Winston Churchill

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...