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Re: Math: Probability [#permalink]
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Man....it must have taken lots of efforts to compile this document.....

kudos to u...


Thanx for sharing this...
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Wow - between Walker and Bunuel, we have Math covered.
Great job!
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Re: Math: Probability [#permalink]
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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]
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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]
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]
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\(C^6_1\) - 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.
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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]
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jasonedward wrote:
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-- something wrong here

(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.


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.
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Re: Math: Probability [#permalink]
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jasonedward wrote:
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.


Donald to receiver either Prograine or Ropecia must be among first two chosen patients and as there are 14 patients then the probability of this is simply 2/14=1/7.
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Re: Math: Probability [#permalink]
walker wrote:
PROBABILITY

This post is a part of [GMAT MATH BOOK]

created by: walker
edited by: bb, Bunuel

--------------------------------------------------------


This topic is included in GMAT ToolKit App (iPhone/iPod Touch)
[read more] [AppStore]



--------------------------------------------------------

A few ways to approach a probability problem

There are a few typical ways that you can use for solving probability questions. Let's consider example, how it is possible to apply different approaches:

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:

3) probability approach: The probability of choosing Bob or Rachel as a first person in committee is 2/8. The probability of choosing Rachel or Bob as a second person when first person is already chosen is 1/7. The probability that the committee includes both Bob and Rachel is.
\(P = \frac{2}{8} * \frac{1}{7} = \frac{2}{56} = \frac{1}{28}\)

4) reversal probability approach: We can choose any first person. Then, if we have Rachel or Bob as first choice, we can choose any other person out of 6 people. If we have neither Rachel nor Bob as first choice, we can choose any person out of remaining 7 people. The probability that the committee includes both Bob and Rachel is.
\(P = 1 - (\frac{2}{8} * \frac{6}{7} + \frac{6}{8} * 1) = \frac{2}{56} = \frac{1}{28}\)


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.

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

2) reversal combinatorial approach: In this example reversal approach is a bit shorter and faster.
\(C^5_1\) - we choose 1 couple out of 5 couples.
\(C^8_1\) - we chose one person out of remaining 8 people.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=1 - \frac{C^5_1*C^8_1}{C^{10}_3}=1 - \frac{5*8}{10*3*4} = \frac{2}{3}\)

3) probability approach:
1st person: \(\frac{10}{10} = 1\) - we choose any person out of 10.
2nd person: \(\frac{8}{9}\) - we choose any person out of 8=10-2(one couple from previous choice)
3rd person: \(\frac{6}{8}\) - we choose any person out of 6=10-4(two couples from previous choices).

\(p = 1*\frac{8}{9}*\frac{6}{8}=\frac{2}{3}\)


Probability tree


Official GMAC Books:

The Official Guide, 12th Edition: DT #4; DT #7; PS #12; PS #67; PS #105; PS #158; PS #174; PS #214; DS #3; DS #107;
The Official Guide, Quantitative 2th Edition: PS #79; PS #160;
The Official Guide, 11th Edition: DT #4; DT #7; PS #10; PS #64; PS #173; PS #217; PS #231; DS #82; DS #114;

Generated from [GMAT ToolKit]


Resources

Probability DS problems: [search]
Probability PS problems: [search]

Walker's post with Combinatorics/probability problems: [Combinatorics/probability Problems]
Bullet's post with probability problems: [Combined Probability Questions]
-------------------------------------------------------

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Can someone explain the reversal probability approach under the first bold face part?

Also please please explain the red text under the married couple example, how did you know to choose 3 couple out of 5? Is it because the question asks for 3 ppl not married? Im completely lost on this example and particularly on the reversal combinatorial approach......
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Re: Math: Probability [#permalink]
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catty2004 wrote:
Can someone explain the reversal probability approach under the first bold face part?

Also please please explain the red text under the married couple example, how did you know to choose 3 couple out of 5? Is it because the question asks for 3 ppl not married? Im completely lost on this example and particularly on the reversal combinatorial approach......


Check similar problems with different approaches:
combination-permutation-problem-couples-98533.html
ps-combinations-94068.html
committee-of-88772.html

Hope it helps.

As for the reverse probability approach, can you please be a bit more specific, what part is confusing for you?
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Re: Math: Probability [#permalink]
fruit wrote:
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?)

Are there any GMAT questions testing this concept?
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Re: Math: Probability [#permalink]
Dear Walker

I was trying your given example- 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? with other method but the answer is coming wrong.

I did 10c1X8c1X6c1/ 10c3. We can select any one random person from 10 then to avoid pairing we remove his/her counterpart so next selection becomes 8c1 and similarly 6c1...but the probability is coming out to be 4. Where am I wrong?

Thanks for your help! :)
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Re: Math: Probability [#permalink]
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ShantnuMathuria wrote:
Dear Walker

I was trying your given example- 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? with other method but the answer is coming wrong.

I did 10c1X8c1X6c1/ 10c3. We can select any one random person from 10 then to avoid pairing we remove his/her counterpart so next selection becomes 8c1 and similarly 6c1...but the probability is coming out to be 4. Where am I wrong?

Thanks for your help! :)


The following post might help: if-4-people-are-selected-from-a-group-of-6-married-couples-99055.html#p764040

Also check similar questions to practice:
a-committee-of-three-people-is-to-be-chosen-from-four-teams-130617.html
if-4-people-are-selected-from-a-group-of-6-married-couples-99055.html
a-committee-of-3-people-is-to-be-chosen-from-four-married-94068.html
if-a-committee-of-3-people-is-to-be-selected-from-among-88772.html
a-comittee-of-three-people-is-to-be-chosen-from-four-married-130475.html
a-committee-of-three-people-is-to-be-chosen-from-4-married-101784.html
a-group-of-10-people-consists-of-3-married-couples-and-113785.html
if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html
given-that-there-are-6-married-couples-if-we-select-only-58640.html
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Re: Math: Probability [#permalink]
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.

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

2) reversal combinatorial approach: In this example reversal approach is a bit shorter and faster.
\(C^5_1\) - we choose 1 couple out of 5 couples.
\(C^8_1\) - we chose one person out of remaining 8 people.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=1 - \frac{C^5_1*C^8_1}{C^{10}_3}=1 - \frac{5*8}{10*3*4} = \frac{2}{3}\)

3) probability approach:
1st person: \(\frac{10}{10} = 1\) - we choose any person out of 10.
2nd person: \(\frac{8}{9}\) - we choose any person out of 8=10-2(one couple from previous choice)
3rd person: \(\frac{6}{8}\) - we choose any person out of 6=10-4(two couples from previous choices).

\(p = 1*\frac{8}{9}*\frac{6}{8}=\frac{2}{3}\)


--------------------------------------------------------


Didn't follow the reverse combinatorial approach here. The question asks the probability that none of the 3 people selected would be married. But the approach is taking 1 married couple and 1 unmarried person. What am I missing?
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Re: Math: Probability [#permalink]
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Aditi10 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.

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

2) reversal combinatorial approach: In this example reversal approach is a bit shorter and faster.
\(C^5_1\) - we choose 1 couple out of 5 couples.
\(C^8_1\) - we chose one person out of remaining 8 people.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=1 - \frac{C^5_1*C^8_1}{C^{10}_3}=1 - \frac{5*8}{10*3*4} = \frac{2}{3}\)

--------------------------------------------------------


Didn't follow the reverse combinatorial approach here. The question asks the probability that none of the 3 people selected would be married. But the approach is taking 1 married couple and 1 unmarried person. What am I missing?


Aditi10,

We can select 3 people out of 10 peoples (5 married couples) in two possible ways:

1. Event A: None of them are married to each other
or
2. Event B: 1 married couple and 1 person.

Note that event A and B are mutually exclusive and exhaustive events.

We know that total probability is always 1. => P(A) + P(B) = 1

In the first approach, we directly compute P(A).

In "reverse combinatorial" approach, we calculate first P(B) and then subtract it from 1 to get P(A).

Hope it helps.
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