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# Math: Probability

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Re: Math: Probability [#permalink]  03 Mar 2010, 22:12
Thanks for both, theory and examples.
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Re: Math: Probability [#permalink]  17 Mar 2010, 11:19
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?
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Re: Math: Probability [#permalink]  18 Mar 2010, 22:06
Expert's post
fruit wrote:
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?

That's right.
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Re: Math: Probability [#permalink]  22 Apr 2010, 08:18
thanks a lot very useful..
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Re: Math: Probability [#permalink]  14 Jul 2010, 18:10
walker wrote:
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.
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Re: Math: Probability [#permalink]  11 Aug 2010, 08:54
walker wrote:
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.

$$p=\frac{C^5_3*(C^2_1)^3}{C^{10}_3}=\frac{10*8}{10*3*4} = \frac{2}{3}$$

can somebody explain this to me...
seems bit complicated....
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Re: Math: Probability [#permalink]  11 Aug 2010, 19:35
Excellent !! great job!
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Re: Math: Probability [#permalink]  31 Aug 2010, 10:20
Amazing job , this is why gmatclub is the best, any one mastering above concepts can mince the GMAT.
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Re: Math: Probability [#permalink]  01 Sep 2010, 21:40
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Re: Math: Probability [#permalink]  12 Sep 2010, 10:42
teriffic job, learnt a lot about using combination for finding probability. GRACIAS...
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Re: Math: Probability [#permalink]  14 Sep 2010, 01:19
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?
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Re: Math: Probability [#permalink]  22 Sep 2010, 13:01
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?
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Re: Math: Probability [#permalink]  22 Sep 2010, 21:43
Expert's post
gurpreetsingh wrote:
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.
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Re: Math: Probability [#permalink]  15 Dec 2010, 07:15
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}
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Re: Math: Probability [#permalink]  15 Dec 2010, 10:04
walker wrote:
$$C^6_1$$ - 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
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Re: Math: Probability [#permalink]  11 Feb 2011, 21:45
Hey guys,

First post..

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.
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Re: Math: Probability [#permalink]  04 Mar 2011, 14:36
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.
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Re: Math: Probability [#permalink]  04 Mar 2011, 18:07
fluke wrote:

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!
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Re: Math: Probability [#permalink]  16 Apr 2011, 13:09
great work Walker..
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Re: Math: Probability [#permalink]  16 Apr 2011, 13:11
Bunuel, really a fan of ur math book..
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Re: Math: Probability   [#permalink] 16 Apr 2011, 13:11

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