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:

There is a 10% chance that it won't snow all winter long [#permalink]

Show Tags

03 Sep 2013, 23:13

2

This post received KUDOS

11

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

44% (02:08) correct
56% (01:11) wrong based on 260 sessions

HideShow timer Statictics

There is a 10% chance that it won't snow all winter long. There is a 20% chance that schools will not be closed all winter long. What is the greatest possible probability that it will snow and schools will be closed during the winter?

(A) 55% (B) 60% (C) 70% (D) 72% (E) 80%

I have seen several explanations on this one and I cannot understand the point of intersection I guess...Has anyone seen similar problems they saw here (specifically with given probabilities) or perhaps I am looking for a parallel (gumballs, etc) to this problem.

There is one rule on dependent probability P(A) + P(B) - P(AB) but clearly this is not the case here. The dice are independent. With colored balls, we could calculate what happens after the balls taken away (when they are not replaced) and it makes sense. With this type of problem, I do not know what to precisely think of dependent probability And what is up with solving probability with overlapping sets or the matrix?

And what exactly happens when we multiply 90% and 80% conceptually? I mean what does that presume and why it is not correct? Basically, a thorough breakdown of this problem would be very helpful.

Please remember that I have seen other explanations, and my questions are based on them. Probably the best of them are: "The greatest possible probability will occur when b is dependent on a. Thus it will be their intersection set. Hence answer = 0.8"

"They're asking for the greatest possible probability. This will occur if whenever it snows, schools close. This is a causation problem. That means schools will close only if it snows, but not all the time it snows. So you don't have to multiply .9*.8, because each time .8 occurs, .9 will be in effect as well."

I do not understand exactly what they mean...Anyway, I hope you see where I am aiming.

Re: There is a 10% chance that it won't snow all winter long [#permalink]

Show Tags

04 Sep 2013, 23:18

7

This post received KUDOS

1

This post was BOOKMARKED

Easily solved with a small table

Here we can infer the values in blue.

So the question asks us what is the maximum for which both C and S are satisfied. The maximum possible value for S is already 80%. For S is 90%. So the maximum for both is 80% (since anything more will no satisfy C i.e. more than 80%).

Re: There is a 10% chance that it won't snow all winter long [#permalink]

Show Tags

05 Sep 2013, 01:30

2

This post received KUDOS

obs23 wrote:

There is a 10% chance that it won't snow all winter long. There is a 20% chance that schools will not be closed all winter long. What is the greatest possible probability that it will snow and schools will be closed during the winter?

(A) 55% (B) 60% (C) 70% (D) 72% (E) 80%

I have seen several explanations on this one and I cannot understand the point of intersection I guess...Has anyone seen similar problems they saw here (specifically with given probabilities) or perhaps I am looking for a parallel (gumballs, etc) to this problem.

There is one rule on dependent probability P(A) + P(B) - P(AB) but clearly this is not the case here. The dice are independent. With colored balls, we could calculate what happens after the balls taken away (when they are not replaced) and it makes sense. With this type of problem, I do not know what to precisely think of dependent probability And what is up with solving probability with overlapping sets or the matrix?

And what exactly happens when we multiply 90% and 80% conceptually? I mean what does that presume and why it is not correct? Basically, a thorough breakdown of this problem would be very helpful.

Please remember that I have seen other explanations, and my questions are based on them. Probably the best of them are: "The greatest possible probability will occur when b is dependent on a. Thus it will be their intersection set. Hence answer = 0.8"

"They're asking for the greatest possible probability. This will occur if whenever it snows, schools close. This is a causation problem. That means schools will close only if it snows, but not all the time it snows. So you don't have to multiply .9*.8, because each time .8 occurs, .9 will be in effect as well."

I do not understand exactly what they mean...Anyway, I hope you see where I am aiming.

We have p (A n B) = p(A) + p(B) - p(A u B)

p(A) is the probability that it will snow p(B) is the probability that the schools will close

p(A n B) will be maximum when p(A u B) is minimum but p(A u B) cannot be less than the higher of the p(A) and p(B) i.e, cannot be less than 0.9.

In the above case B is totally dependent on A. i.,e schools close whenever and only when it snows. So the p(AUB) effectively becomes p(A).

The answer is p(A n B) = 0.9+0.8-0.9= 0.8

In other words the greatest probability is the probability of B. _________________

Re: There is a 10% chance that it won't snow all winter long [#permalink]

Show Tags

05 Sep 2013, 06:03

Helpful guys. Appreciate your efforts. Sravna just to clarify:

We have p (A n B) = p(A) + p(B) - p(A u B)

p(A) is the probability that it will snow p(B) is the probability that the schools will close

Quote:

p(A n B) will be maximum when p(A u B) is minimum

You mean when p(A u B) maximized, don't you?

In other words, the P(ANB) is maximized when P(AORB) is maximized and this maximization happens only when B is completely dependent on A, becoming A and P(AORB)=P(B). Seem to make sense now. Hope I won't be trapped next time. _________________

There are times when I do not mind kudos...I do enjoy giving some for help

Re: There is a 10% chance that it won't snow all winter long [#permalink]

Show Tags

05 Sep 2013, 06:09

adg142000 wrote:

hi obs,

there are 2 events 1) p(A) (probability it will snow all longer)=0.1 2) p(B)(probability the school will remain close all winter longer)= 1- 0.2 = 0.8

now if you go by rule for joint sets p(A u B) = p(A)+p(B) - p(AandB) therefore p(AandB)= 0.1 + 0.8 - p(A u B)

now you can just think that p(AandB) could only be max when B is subset of A thereby P(A u B) would be 0.1

hence max p(AandB) would be 0.8,

I hope i made sense

It looks like you came from another angle, but it seems to me there are some mix ups. P(A) - will not snow all winter long? Anyhow, was wondering if what you meant was similar to what Sravna was saying? Thanks. _________________

There are times when I do not mind kudos...I do enjoy giving some for help

Re: There is a 10% chance that it won't snow all winter long [#permalink]

Show Tags

05 Sep 2013, 06:32

obs23 wrote:

adg142000 wrote:

hi obs,

there are 2 events 1) p(A) (probability it will snow all longer)=0.1 2) p(B)(probability the school will remain close all winter longer)= 1- 0.2 = 0.8

now if you go by rule for joint sets p(A u B) = p(A)+p(B) - p(AandB) therefore p(AandB)= 0.1 + 0.8 - p(A u B)

now you can just think that p(AandB) could only be max when B is subset of A thereby P(A u B) would be 0.1

hence max p(AandB) would be 0.8,

I hope i made sense

It looks like you came from another angle, but it seems to me there are some mix ups. P(A) - will not snow all winter long? Anyhow, was wondering if what you meant was similar to what Sravna was saying? Thanks.

yeah i just missed that ,, you are right p(A) would be 0.9, blunder on my part ,, _________________

Re: There is a 10% chance that it won't snow all winter long [#permalink]

Show Tags

05 Sep 2013, 08:47

obs23 wrote:

Helpful guys. Appreciate your efforts. Sravna just to clarify:

We have p (A n B) = p(A) + p(B) - p(A u B)

p(A) is the probability that it will snow p(B) is the probability that the schools will close

Quote:

p(A n B) will be maximum when p(A u B) is minimum

You mean when p(A u B) maximized, don't you?

In other words, the P(ANB) is maximized when P(AORB) is maximized and this maximization happens only when B is completely dependent on A, becoming A and P(AORB)=P(B). Seem to make sense now. Hope I won't be trapped next time.

Hi,

I meant p(A u B) is minimized i.e., the probability either A or B occurring is minimum. This minimum is equal to the higher of p(A) and p(B) because it cannot be less than that. p(A) is the higher probability. So p(AUB) is equal to p(A).

So p(A n B) = p(A) + p(B) - p(A u B) becomes p(A n B) = p(A) + p(B) -p(A) p(A n B) = p(B)= 0.8 _________________

Re: There is a 10% chance that it won't snow all winter long [#permalink]

Show Tags

06 Sep 2013, 23:27

1

This post received KUDOS

Expert's post

obs23 wrote:

There is a 10% chance that it won't snow all winter long. There is a 20% chance that schools will not be closed all winter long. What is the greatest possible probability that it will snow and schools will be closed during the winter?

(A) 55% (B) 60% (C) 70% (D) 72% (E) 80%

I have seen several explanations on this one and I cannot understand the point of intersection I guess...Has anyone seen similar problems they saw here (specifically with given probabilities) or perhaps I am looking for a parallel (gumballs, etc) to this problem.

There is one rule on dependent probability P(A) + P(B) - P(AB) but clearly this is not the case here. The dice are independent. With colored balls, we could calculate what happens after the balls taken away (when they are not replaced) and it makes sense. With this type of problem, I do not know what to precisely think of dependent probability And what is up with solving probability with overlapping sets or the matrix?

And what exactly happens when we multiply 90% and 80% conceptually? I mean what does that presume and why it is not correct? Basically, a thorough breakdown of this problem would be very helpful.

Please remember that I have seen other explanations, and my questions are based on them. Probably the best of them are: "The greatest possible probability will occur when b is dependent on a. Thus it will be their intersection set. Hence answer = 0.8"

"They're asking for the greatest possible probability. This will occur if whenever it snows, schools close. This is a causation problem. That means schools will close only if it snows, but not all the time it snows. So you don't have to multiply .9*.8, because each time .8 occurs, .9 will be in effect as well."

I do not understand exactly what they mean...Anyway, I hope you see where I am aiming.

Ok, let's try to understand the question stem:

There is a 10% chance that it won't snow all winter long. (by the way, there is some ambiguity in this statement) implies there is a 90% chance that it will snow all winter long i.e. 9 out of 10 winters it will snow all winter long.

There is a 20% chance that schools will not be closed all winter long. implies there is an 80% chance that schools will be closed all winter long i.e. 8 out of 10 winters school will be closed all winter long.

What is the greatest possible probability that it will snow and schools will be closed during the winter?

Attachment:

Ques3.jpg [ 9.22 KiB | Viewed 3508 times ]

What is the greatest possible overlap between the two? The 8 winters when the school is closed, it should snow all winter long too. That's when the overlap will be maximum. So probability is 0.8 _________________

Re: There is a 10% chance that it won't snow all winter long [#permalink]

Show Tags

08 Sep 2013, 02:40

1

This post received KUDOS

Hi Karishma

While i get the logic of your answer, i always understood that probability of event A and Probability of event B occurring is P(A)*P(B). So here is applied the same and got the answer as 0.9*0.8 = 0.72. Whats wrong with it? Please help to explain. _________________

“Confidence comes not from always being right but from not fearing to be wrong.”

Re: There is a 10% chance that it won't snow all winter long [#permalink]

Show Tags

08 Oct 2014, 10:21

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: There is a 10% chance that it won't snow all winter long [#permalink]

Show Tags

24 Jul 2015, 10:48

ankur1901 wrote:

Hi Karishma

While i get the logic of your answer, i always understood that probability of event A and Probability of event B occurring is P(A)*P(B). So here is applied the same and got the answer as 0.9*0.8 = 0.72. Whats wrong with it? Please help to explain.

There is a 10% chance that it won't snow all winter long [#permalink]

Show Tags

24 Jul 2015, 19:30

obs23 wrote:

There is a 10% chance that it won't snow all winter long. There is a 20% chance that schools will not be closed all winter long. What is the greatest possible probability that it will snow and schools will be closed during the winter?

(A) 55% (B) 60% (C) 70% (D) 72% (E) 80%

I have seen several explanations on this one and I cannot understand the point of intersection I guess...Has anyone seen similar problems they saw here (specifically with given probabilities) or perhaps I am looking for a parallel (gumballs, etc) to this problem.

There is one rule on dependent probability P(A) + P(B) - P(AB) but clearly this is not the case here. The dice are independent. With colored balls, we could calculate what happens after the balls taken away (when they are not replaced) and it makes sense. With this type of problem, I do not know what to precisely think of dependent probability And what is up with solving probability with overlapping sets or the matrix?

And what exactly happens when we multiply 90% and 80% conceptually? I mean what does that presume and why it is not correct? Basically, a thorough breakdown of this problem would be very helpful.

Please remember that I have seen other explanations, and my questions are based on them. Probably the best of them are: "The greatest possible probability will occur when b is dependent on a. Thus it will be their intersection set. Hence answer = 0.8"

"They're asking for the greatest possible probability. This will occur if whenever it snows, schools close. This is a causation problem. That means schools will close only if it snows, but not all the time it snows. So you don't have to multiply .9*.8, because each time .8 occurs, .9 will be in effect as well."

I do not understand exactly what they mean...Anyway, I hope you see where I am aiming.

The greatest probability translates as "take the range of probabilities of these two events occurring under different circumstances, and choose the greatest extreme."

If event B is fully dependent on event A, then the probability is equal to the smallest probability of either event A or B, 0.8.

If event B is not fully dependent on event A, then the maximum we can know for sure is 0.72, or event A * event B.

So, the range of possibilities is 0.72 to 0.8. The greatest possibility is 0.8 or 80%.

There is a 10% chance that it won't snow all winter long [#permalink]

Show Tags

30 Dec 2015, 22:14

3

This post received KUDOS

Expert's post

dina98 wrote:

Could someone please explain how P(A intersection B) is calculated differently from P(A U B)? Thanks

A intersection B is whatever is common between the two sets.

Either you will be given its value or if they are independent events, then P(A intersection B) = P(A) * P(B).

For mutually exclusive events, P(A intersection B) = 0. Mutually exclusive events are those which cannot happen at the same time such as "Getting heads on flipping a coin" and "Getting tails on flipping a coin".

When there is some dependence in the events, the maximum value the intersection can have is the lower of the two probabilities. Say P(A) = 0.4 and P(B) = 0.7. The maximum probability of intersection can be 0.4 because P(A) = 0.4. The minimum value of P(A intersection B) will be 0.1 since probability cannot exceed 1 so P(A U B) is maximum 1. 1 = 0.4 + 0.7 - P(A intersection B) P(A intersection B) = 0.1 (at least)

Actual value of P(A intersection B) will lie somewhere between 0.1 and 0.4 (inclusive)

A union B is the sum of whatever is common between them and the elements which are contained in one set only. P(A U B) = P(A) + P(B) - P(A intersection B) _________________

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...