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

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There is a 10% chance that it won't snow all winter long [#permalink] New post 03 Sep 2013, 22:13
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35% (02:12) correct 64% (00:50) wrong based on 81 sessions
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.
[Reveal] Spoiler: OA

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Re: There is a 10% chance that it won't snow all winter long [#permalink] New post 04 Sep 2013, 20:53
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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] New post 04 Sep 2013, 22:18
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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%).

Hope it helps
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Re: There is a 10% chance that it won't snow all winter long [#permalink] New post 05 Sep 2013, 00:30
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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] New post 05 Sep 2013, 05:03
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.
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Re: There is a 10% chance that it won't snow all winter long [#permalink] New post 05 Sep 2013, 05:09
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.
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Re: There is a 10% chance that it won't snow all winter long [#permalink] New post 05 Sep 2013, 05:32
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 ,,
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Re: There is a 10% chance that it won't snow all winter long [#permalink] New post 05 Sep 2013, 07:47
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
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Re: There is a 10% chance that it won't snow all winter long [#permalink] New post 06 Sep 2013, 22:27
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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.


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
Ques3.jpg [ 9.22 KiB | Viewed 690 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
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Re: There is a 10% chance that it won't snow all winter long [#permalink] New post 08 Sep 2013, 01:40
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] 08 Sep 2013, 01:40
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