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Hey guys..Please help me out here!! Probability

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Hey guys..Please help me out here!! Probability

by dddanny2006 » Sun Nov 24, 2013 2:26 am
If the probabilities are respectively 0.09,0.15,.21 and 0.23 that a person purchasing a new automobile will choose the color green,white,red or blue,what is the probability that a given buyer will purchase a new automobile that comes in one of those colors.

Answer 0.68---------------Simple method

Method 1

Prob(of buying one of those colours)= 0.09 + (.91*0.15)+(0.91*0.85*0.21)+(0.91*0.85*0.79*0.23)=0.52

But the books says 0.68,why am I wrong?

No problem,Ill do the second method Method 2

Prob(of choosing one of those colours)=1-prob(of not choosing one of those colours)

Prob(of not choosing one of those colours)= 0.91*0.85*0.79*0.77=0.48

There 1-P=0.52

Both Method 1 and 2 support my answer.Why is the answer 0.68.?


Thanks

Dan
Last edited by dddanny2006 on Sun Nov 24, 2013 4:45 am, edited 1 time in total.
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by GMATGuruNY » Sun Nov 24, 2013 3:59 am
dddanny2006 wrote:If the probabilities are respectively 0.09,0.15,.21 and 0.23 that a person purchasing a new automobile will choose the color green,white,red or blue,what is the probability that a given buyer will purchase a new automobile that comes in one of those colors.
And means MULTIPLY:
P(A and B) = P(A) * P(B).

Or means ADD:
P(A or B) = P(A) + P(B).

The problem above is an OR problem.
The question stem asks for the probability that the person chooses green OR white OR red OR blue.
Thus, we ADD the probabilities:
(0.09) + (0.15) + (0.21) + (0.23) = 0.68.
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by dddanny2006 » Sun Nov 24, 2013 4:32 am
Thanks for that.Why is my method wrong?What answer does my method relate to?I have attached a problem below.Whats the difference between the above problem and the attached one.My argument is that he either buys a green car,or doesnt buy a green but goes for white and so on etc etc.Please clear my confusion.
GMATGuruNY wrote:
dddanny2006 wrote:If the probabilities are respectively 0.09,0.15,.21 and 0.23 that a person purchasing a new automobile will choose the color green,white,red or blue,what is the probability that a given buyer will purchase a new automobile that comes in one of those colors.
And means MULTIPLY:
P(A and B) = P(A) * P(B).

Or means ADD:
P(A or B) = P(A) + P(B).

The problem above is an OR problem.
The question stem asks for the probability that the person chooses green OR white OR red OR blue.
Thus, we ADD the probabilities:
(0.09) + (0.15) + (0.21) + (0.23) = 0.68.
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by kackerarnav » Sun Nov 24, 2013 6:11 am
Hi Dan,

The core difference in the questions lies the sequential nature of the probabilities of the second, as opposed to the scenario in the first question.

The probabilities given in Q1 are entirely representative of the chances of picking each of those colours and do not need to be adjusted for him progressing 'through a list' of those colours. Hence, by multiplying the probability of picking green with the probability of not picking red, you are introducing the fact that he doesn't get a shot at choosing green unless he's eliminated red first - which is not the case.

In the second question, he does not get a shot at checking out whether the second station is playing a song he likes, unless he's observed and eliminated the first station. So, this is in fact sequential.

As far as the second method you tried out for the first question goes, you multiplied all the probabilities together, which gave you the probability of him buying cars of ALL these colours. Subtracting that from one gave you the probability of him not buying cars of all these colours. This is neither what's asked for nor what the probabilities represent.

You need the probability of buying one of these colours:

P (colour A) OR P (colour B) OR P (colour C) OR P (colour D)

The OR probability function is represented by addition. Hence, adding these wholly representative probabilities together will give you the answer you desire.

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by dddanny2006 » Sun Nov 24, 2013 7:47 am
Well I did understand the first part.But when we use the 1-p method we get to know that atleast one of those cars is bought if not all.It doesnt only mean that all cars are bought,but also atleast one car is bought.My doubt still remains.
kackerarnav wrote:Hi Dan,

The core difference in the questions lies the sequential nature of the probabilities of the second, as opposed to the scenario in the first question.

The probabilities given in Q1 are entirely representative of the chances of picking each of those colours and do not need to be adjusted for him progressing 'through a list' of those colours. Hence, by multiplying the probability of picking green with the probability of not picking red, you are introducing the fact that he doesn't get a shot at choosing green unless he's eliminated red first - which is not the case.

In the second question, he does not get a shot at checking out whether the second station is playing a song he likes, unless he's observed and eliminated the first station. So, this is in fact sequential.

As far as the second method you tried out for the first question goes, you multiplied all the probabilities together, which gave you the probability of him buying cars of ALL these colours. Subtracting that from one gave you the probability of him not buying cars of all these colours. This is neither what's asked for nor what the probabilities represent.

You need the probability of buying one of these colours:

P (colour A) OR P (colour B) OR P (colour C) OR P (colour D)

The OR probability function is represented by addition. Hence, adding these wholly representative probabilities together will give you the answer you desire.

Best regards,

Arnav
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by Mike@Magoosh » Mon Nov 25, 2013 10:44 am
dddanny2006 wrote:Well I did understand the first part.But when we use the 1-p method we get to know that atleast one of those cars is bought if not all.It doesnt only mean that all cars are bought,but also atleast one car is bought.My doubt still remains.
Dear dddanny2006,
I am responding to your private message. You are using rules and solution methods in probability without any understanding of their basis. I strongly suggest finding not a single-page summary, but a serious textbook chapter on probability and reading through it carefully. You need to absorb the underlying ideas of probability. The very worst way to approach probability is a formula-based approach. Probability involves a whole worldview, and until you integrate this worldview and are asking the right questions about a situation, you will just be picking formulas more or less at random.

Two absolutely essential ideas in this worldview are the ideas of
(1) mutually exclusive
(2) independent
You cannot begin to apply any of the probability rules until you understand the problem through the perspective of these ideas.

In this scenario, the colors are mutually exclusive --- a car doesn't come in multiple colors, so the entire car will be only green, or only white, or etc. If the car is one color, this excludes all the others. Notice, also, this is what I would call a side-by-side situation, as most probability scenarios are, not a sequential situation. Your approach at the top --- "if not this, then that; if not this and this, then that" --- would be appropriate only for a sequential situation. In order to apply a sequential solution, the problem would have to specify a sequence of events, not merely a side-by-side choice.

There is no fixed order to the colors here, and no sequential choice. We can imagine the prospective customer walking in the car dealership, and on the showroom floor there could be four cars side-by-side --- a green car, a white car, a red card, and a blue car. The customer sees all four at once, and chooses one from the four. There is no reason a person needs to "choose against" any color before choosing the color she wants. The person sees all the colors, and in one choice, chooses one.

When events A & B & C & D are mutually exclusive, we can use the very simple "OR" rule:
P(A or B or C or D) = P(A) + P(B) + P(C) + P(D)

All we have to do here is add the four numbers given. We know this only by thinking carefully about the nature of the situation.

Does all this make sense?
Mike :-)
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by dddanny2006 » Mon Nov 25, 2013 10:52 am
Brilliant..Why doesnt the 1-P method work here?What does the 1-p method suggest?Is it TRUE that 1-P method should not be used for Side by side problems?
Mike@Magoosh wrote:
dddanny2006 wrote:Well I did understand the first part.But when we use the 1-p method we get to know that atleast one of those cars is bought if not all.It doesnt only mean that all cars are bought,but also atleast one car is bought.My doubt still remains.
Dear dddanny2006,
I am responding to your private message. You are using rules and solution methods in probability without any understanding of their basis. I strongly suggest finding not a single-page summary, but a serious textbook chapter on probability and reading through it carefully. You need to absorb the underlying ideas of probability. The very worst way to approach probability is a formula-based approach. Probability involves a whole worldview, and until you integrate this worldview and are asking the right questions about a situation, you will just be picking formulas more or less at random.

Two absolutely essential ideas in this worldview are the ideas of
(1) mutually exclusive
(2) independent
You cannot begin to apply any of the probability rules until you understand the problem through the perspective of these ideas.

In this scenario, the colors are mutually exclusive --- a car doesn't come in multiple colors, so the entire car will be only green, or only white, or etc. If the car is one color, this excludes all the others. Notice, also, this is what I would call a side-by-side situation, as most probability scenarios are, not a sequential situation. Your approach at the top --- "if not this, then that; if not this and this, then that" --- would be appropriate only for a sequential situation. In order to apply a sequential solution, the problem would have to specify a sequence of events, not merely a side-by-side choice.

There is no fixed order to the colors here, and no sequential choice. We can imagine the prospective customer walking in the car dealership, and on the showroom floor there could be four cars side-by-side --- a green car, a white car, a red card, and a blue car. The customer sees all four at once, and chooses one from the four. There is no reason a person needs to "choose against" any color before choosing the color she wants. The person sees all the colors, and in one choice, chooses one.

When events A & B & C & D are mutually exclusive, we can use the very simple "OR" rule:
P(A or B or C or D) = P(A) + P(B) + P(C) + P(D)

All we have to do here is add the four numbers given. We know this only by thinking carefully about the nature of the situation.

Does all this make sense?
Mike :-)
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by Mike@Magoosh » Mon Nov 25, 2013 12:07 pm
dddanny2006 wrote:Brilliant..Why doesn't the 1-P method work here? What does the 1-P method suggest?Is it TRUE that 1-P method should not be used for Side by side problems?
Danny
There is no such thing as the "1-P method." Nobody who understand probability theory uses that terminology. Your very question betrays a much wider gap of understanding that merely the content of the question. You are grabbing context specific solutions to various problems and reifying them as "methods". All of probability is deeply contextual, and you are trying to create "methods" that work apart fro this context.
Read these two blogs, carefully and thoroughly:
https://magoosh.com/gmat/2012/gmat-math- ... ity-rules/
https://magoosh.com/gmat/2012/gmat-math- ... -question/

Then, if you haven't answered your own question in doing so, ask me your question again, using the proper terminology. For your sake, I am going to force you to think more carefully about the ideas of probability, rather than merely address a superficial answer to the level from which you ask. In other words, I am trying to educate you, not merely on the explicit question you are asking, but on everything I can see that you don't yet understand.

Does this make sense?
Mike :-)
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by dddanny2006 » Mon Nov 25, 2013 12:11 pm
Yes,Im finding probability very hard.Thanks for the links.Can you suggest any book that will have comprehensive coverage so that I dont miss out on anything?
I have a bad habit of questioning every solution as to "why cant it be done this way or like that problem".

Mike@Magoosh wrote:
dddanny2006 wrote:Brilliant..Why doesn't the 1-P method work here? What does the 1-P method suggest?Is it TRUE that 1-P method should not be used for Side by side problems?
Danny
There is no such thing as the "1-P method." Nobody who understand probability theory uses that terminology. Your very question betrays a much wider gap of understanding that merely the content of the question. You are grabbing context specific solutions to various problems and reifying them as "methods". All of probability is deeply contextual, and you are trying to create "methods" that work apart fro this context.
Read these two blogs, carefully and thoroughly:
https://magoosh.com/gmat/2012/gmat-math- ... ity-rules/
https://magoosh.com/gmat/2012/gmat-math- ... -question/

Then, if you haven't answered your own question in doing so, ask me your question again, using the proper terminology. For your sake, I am going to force you to think more carefully about the ideas of probability, rather than merely address a superficial answer to the level from which you ask. In other words, I am trying to educate you, not merely on the explicit question you are asking, but on everything I can see that you don't yet understand.

Does this make sense?
Mike :-)
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by Mike@Magoosh » Mon Nov 25, 2013 12:56 pm
dddanny2006 wrote:Yes,Im finding probability very hard.Thanks for the links.Can you suggest any book that will have comprehensive coverage so that I dont miss out on anything?
I have a bad habit of questioning every solution as to "why cant it be done this way or like that problem".
Danny,
I am going to recommend the deluxe treatment for you. There's a guy, David S Moore, who has written a series of statistics text books. If you go to amazon.com and search for "Moore Statistics", you will get hundreds of results. Basically, a used copy of any one of them will do, anything with a title something like "the practice of statistics" or "introduction to the practice of statistics", any edition --- you might want to check at a local used bookstore or wherever used textbooks are sold; also, there are some used version on amazon that sell for less than $5. One section in any good stats book is a thorough treatment of probability, as least as much as you need for the GMAT ---- typically, it consists of a few chapters. It's more or less the same information recycled in all these different books with Moore's name on it.
Incidentally, Chapter 1-2, on introductory statistics, all information up to but not including the "normal distribution", would also tell you everything you need to know about this topic for the GMAT. If you tell me what edition you have gotten, I should be able to tell you exactly what chapters to read. If you have questions about probability theory from that book, post them here, and I will be happy to answer your questions.
Does all this make sense?
Mike :-)
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by dddanny2006 » Mon Nov 25, 2013 2:05 pm
Hey Mike.

I have the book with me.Can you please tell me what chapters to go through?
Mike@Magoosh wrote:
dddanny2006 wrote:Yes,Im finding probability very hard.Thanks for the links.Can you suggest any book that will have comprehensive coverage so that I dont miss out on anything?
I have a bad habit of questioning every solution as to "why cant it be done this way or like that problem".
Danny,
I am going to recommend the deluxe treatment for you. There's a guy, David S Moore, who has written a series of statistics text books. If you go to amazon.com and search for "Moore Statistics", you will get hundreds of results. Basically, a used copy of any one of them will do, anything with a title something like "the practice of statistics" or "introduction to the practice of statistics", any edition --- you might want to check at a local used bookstore or wherever used textbooks are sold; also, there are some used version on amazon that sell for less than $5. One section in any good stats book is a thorough treatment of probability, as least as much as you need for the GMAT ---- typically, it consists of a few chapters. It's more or less the same information recycled in all these different books with Moore's name on it.
Incidentally, Chapter 1-2, on introductory statistics, all information up to but not including the "normal distribution", would also tell you everything you need to know about this topic for the GMAT. If you tell me what edition you have gotten, I should be able to tell you exactly what chapters to read. If you have questions about probability theory from that book, post them here, and I will be happy to answer your questions.
Does all this make sense?
Mike :-)
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by Mike@Magoosh » Mon Nov 25, 2013 2:18 pm
dddanny2006 wrote:Hey Mike.

I have the book with me.Can you please tell me what chapters to go through?
Danny,
Another thing we need to work on is precision. Math is all about precision. The words "the book" tells me zero. Tell me the exact title, the authors, the edition, and the ISBN.
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by dddanny2006 » Mon Nov 25, 2013 2:24 pm
Cool
Introduction to the practice of Statistics by David S Moore,Mcabe and Craig 6th Edition.I have attached the ISBN.




Mike@Magoosh wrote:
dddanny2006 wrote:Hey Mike.

I have the book with me.Can you please tell me what chapters to go through?
Danny,
Another thing we need to work on is precision. Math is all about precision. The words "the book" tells me zero. Tell me the exact title, the authors, the edition, and the ISBN.
Mike
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