There is a 10% chance that it won't snow all winter long : Problem Solving (PS)

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

05 Sep 2013, 01:30

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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.

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

30 Dec 2015, 22:14

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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)

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

13 Jun 2020, 04:57

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varotkorn wrote:

If the question were What is the MINIMUM probability that it will snow and schools will be closed during the winter?

Would the answer be 0.7 (using matrix/grid) or 0.72 (by assuming independent)?

You multiply probabilities when you have two unrelated ("independent") events, and want to know the probability both events will happen. Multiplication has nothing to do with finding "minimum" or "maximum" probabilities.

If you're asked to find the "minimum" or "maximum" probability two events occur, the events really must be *dependent* (related in some way). If they were independent, there's a single numerical answer to the question "what is the probability both events occur?" so there'd be no reason to ask for a "minimum" or "maximum". For example, it wouldn't make sense to ask "if you flip two coins, what is the minimum probability both coins turn up Heads?" because there is an exact answer to that question: it's 1/4. There's no "minimum" or "maximum" answer.

So if you're asked to find a minimum or maximum probability of two events occurring, the events will be dependent, and then you always need to use Venn diagram (or matrix, if you prefer) methods to answer. You'll just want to try to maximize or minimize the overlap in your Venn diagram, using the fact that the probabilities of all of the events must sum to 1. As you found, the minimum probability in this question is 0.7 (which you find by assuming at least one event must happen).

If you do multiply 0.9 and 0.8 in this question, all you are doing is finding one possible answer to the question "what is the probability both events occur?", the answer when the two events happen to be independent. The answer you find, 0.72, won't be the minimum or the maximum -- it will be somewhere in between the min and max.

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

04 Sep 2013, 21:53

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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

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

08 Sep 2013, 02:40

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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.

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

05 Oct 2015, 00:38

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Basically, the probability is 0.72 when both of the events are mutually exclusive. When they are dependent you got to use the equation

P(AUB) = P(A) + P(B) - P(A intersection B)

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

05 Sep 2013, 08:47

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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

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

24 Jul 2015, 19:30

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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%.

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

01 Dec 2017, 00:46

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Max probability of both event A & B = least value of A or B
this is because P(A& B) = P(A)*P(B) , if the events are independent.
P(A& B) = P(A)*P(B/A) , if the events are independent.
now since the max value of probability for any event P(B) or P(B/A)= 1, hence max probability of both events = min of probability of individual events.
Here probability of snow P(A) =0.9
Probability of school closing, P(B) =0.8
Hence probability of their intersection =0.8.
Hope it is clear.
Please provide Kudos if u like the solution.

Sent from my 2014818 using GMAT Club Forum mobile app

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

05 Sep 2013, 06:03

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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.

B

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

30 Dec 2015, 12:40

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

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

13 Jun 2020, 00:46

Dear IanStewart GMATGuruNY VeritasKarishma Bunuel,

If the question were What is the MINIMUM probability that it will snow and schools will be closed during the winter?

Would the answer be 0.7 (using matrix/grid) or 0.72 (by assuming independent)?

P

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

29 Aug 2023, 18:07

Hey karishma - thanks for your explanation below.
However, why would we not multiply the two probabilities? I.e. 0.8 * 0.9 = 0.72 = 72% ? I marked 72% for both events to happen together. I did not see it from the sens you're seeing it. Can you help me understand a bit more about it? Thanks. Also as per @ianstewart's response above - how do we conclude the events in question are dependent or independent? Snowing and closing of schools are independent activities right? My thinking is that if there are 10 marbles (2 red 8 green) in a bag, and I want to draw out 2 red marbles one by one then my probability is 2/10*1/9. So such an event is dependent but here there is nothing as such which proves them dependent

KarishmaB wrote:

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

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

B

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

29 Nov 2023, 15:39

snow no snow Total
school 10 10 20
No school 80 0 80
Total 90 10 100

To maximize 'no school and snow', I used it as zero, and then it shows as 80%.

Posted from my mobile device

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

29 Nov 2023, 15:39

GMAT Problem Solving (PS) Questions

There is a 10% chance that it won't snow all winter long