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# I'm going to try to collect some P and C type questions and

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I'm going to try to collect some P and C type questions and [#permalink]

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09 Mar 2005, 08:41
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I'm going to try to collect some P and C type questions and approaches in this thread for easy reference for everybody. Please feel free to discuss and add to the thread.

Last edited by HongHu on 09 Mar 2005, 09:33, edited 1 time in total.
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09 Mar 2005, 08:54
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Binomial Distribution

In each individual test, the probability of A happening is p and not happening is 1-p. What is the probability of A happening exactly k times in n repeated tests?

Formula:
C(n,k) * p^k * (1-p) ^ n-k

Example:
When a coin has tossed, it will show a head at 50% probability and a tail at 50% probability. What is the probability of getting two heads among four tosses?
C(4, 2) * (1/2) ^2 * (1/2) ^2 = 6/16 = 3/8

Example:
The probability of raining is 0.3 and not raining is 0.7. What is the probability of getting three days of rains among seven days?
C(7, 3)*0.3^3*0.7^4

Example:
The probability of a new born baby is a boy is 50% and that of a new born baby is a girl is 50%. Ten mothers are going to give birth to ten babies this morning. What is the probability that at least two babies are boys?

Probability that at least two babies are boys
= Probability that two babies are boys + probability that three babies are boys + ... + Probability that ten babies are boys
(also) = 1- Probability that non are boys - probability that only one is a boy
Choose the easier route
P=1-C(10,0)*0.5^0*0.5^10-C(10,1)*0.5^1*0.5^9
=1-0.5^10-10*0.5^10
=1-11*0.5^10
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09 Mar 2005, 09:29
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Repeated Outcomes

If there're only k possible outcomes for each object, total possible outcomes for n objects is k^n.

Explanation: For each object, there are k outcomes. So the total number of outcomes would be k*k*k ...*k, for n times.

Example:

Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there are some fruits in the basket, how many different choices does she have? (Assuming the fruits are all different from each other.)
Total number of fruits = 9
For each fruit there are 2 possible outcomes: included, not included
Total outcome = 2^9-1
(The minus one is to take out the one possible outcome where nothing is in the basket.)

Example:

Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there is at least one of each kind, how many different choices does she have? (Assuming the fruits are all different from each other.)
Total outcome = (2^3-1)*(2^4-1)*(2^2-1)

Example:

There are three secretaries who work for three departments. If each of the three departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?

For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C
Total outcome: 3^3=27
Total outcome that three secretary are assigned at least one report: P(3,3)=3!=6
Probability = 6/27=2/9

Example:

There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?

For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C
Total outcome: 3^4=81
Total outcome that three secretary are assigned at least one report: We need to choose one secretary C(3,1) to be assigned of two reports C(4,2) and then the rest two secretary each to be assigned of one report P(2,2). C(3,1)*C(4,2)*P(2,2)=36
Probability = 36/81=4/9

Question: in the above example, why isn't total outcome = 4^3?
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09 Oct 2006, 04:55
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permutations theoretical concept (sorry first time the file hasn't attached)
Attachments

Permutation.doc [78.5 KiB]

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24 Feb 2007, 08:15
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HongHu wrote:
Letters and Words

A common GMAT P&C problem type is letter problems. You would be asked to take letters from an existing word to form new words.

Example:
Four letters are taken at random from the word AUSTRALIA. What is the probability to have two vowels and two consonants?

Solution:

Vowels: A, U, A, I, A.
Consonants: S, T, R, L

This is a case of "without replacement".
Total outcomes: C(9,4)=9*8*7*6/4!=126
Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60
Probability=60/126=10/21

Can anyone please explain why we can't count the probability like this?:

5/9 x 4/8 x 4/7 x 3/6 (VxVxCxC) = 5/63

The way you are doing it, you've given it a specific order.

For example, if there are 2 red balls and 2 white balls, what is the probability of getting 1 white ball and 1 red ball out in the first two draws? It's C(2,1)*C(2,1)/C(4,2) = 2/3.
You could write it out to verify:
R1R2
R1W1
R1W2
R2R1
R2W1
R2W2
W1R1
W1R2
W1W2
W2R1
W2R2
W2W1

But using your method you will have 2/4*2/3=1/3. Your problem is you have specified that you will draw a white ball first, and then red ball second. However there is also the possibility of drawing a red ball first, and then a white ball second. So you need to add another 1/3 to your result to get 2/3.
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09 Mar 2005, 10:33
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Example:

Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there are some fruits in the basket, how many different choices does she have? (Assuming the fruits are all different from each other.)
Total number of fruits = 9
For each fruit there are 2 possible outcomes: included, not included
Total outcome = 2^9-1
(The minus one is to take out the one possible outcome where nothing is in the basket.)

Alternative Solution (Not Recommended because too long):

((3c0+3c1+3c2+3c3)*(4c0+4c1+4c2+4c3+4c4)*(2c0+2c1+2c2))-1
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09 Mar 2005, 15:57
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Letters and Words

A common GMAT P&C problem type is letter problems. You would be asked to take letters from an existing word to form new words.

Example:
Four letters are taken at random from the word AUSTRALIA. What is the probability to have two vowels and two consonants?

Solution:

Vowels: A, U, A, I, A.
Consonants: S, T, R, L

This is a case of "without replacement".
Total outcomes: C(9,4)=9*8*7*6/4!=126
Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60
Probability=60/126=10/21

Example:
From the word AUSTRALIA, a letter is taken at random and put back. Then a second letter is taken and put back. This process is repeated for four letters. What would be the possibility that two vowels and two consonants have been chosen?

This is a case of "with replacement":
Total outcomes: 9^4
Outcomes with two vowels and two consonants: C(4,2)*5^2*4^2
You can calculate the probability from here.

Note that there is a special thing with letter problems, ie. the repeating letters. In the above examples the repeating letters don't matter. However they would matter if the question is asked differently.

Example:
There is a word AUSTRALIA. Four letters are taken at random. How many different words can be formed that have two vowels and two consonants?

Solution:
Vowels: A, U, A, I, A. Distinguish vowles: A, U, I.
Consonants: S, T, R, L

Outcomes with two vowels and two consonants:
1) The two vowels are different: C(3,2)*C(4,2)
2) The two vowels are the same: C(1,1)*C(4,2)

After you've got the four letters you need to order them to get different outcomes. So the first case (with different vowels) would be C(3,2)*C(4,2)*P(4,4)=432, and the second case (with same vowels) would be C(1,1)*C(4,2)*P(4,4)/2=72. The final outcome would be 504.

Last edited by HongHu on 25 Apr 2011, 05:50, edited 2 times in total.
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11 Mar 2005, 12:28
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Very good question. When we do 9^4, 5^2 and 4^2 we've already taken count for the cases of different orders. For example, for the two vowels AU and UA are counted as two occasions. We don't need to order those four letters again. In other words we don't need to multiply 5^2*4^2 by 4!. However, what we have not counted, is that the two vowels can be placed onto different space. For example, for each set of two vowels and two consonants, say AU and ST, we could do different placements: AUST, ASUT, or ASTU, etc. They are all different words that need to be counted. Therefore we need to pick two spaces for the two vowels out of the four spaces to account for this. That's why we multiply 5^2*4^2 by C(4,2). The rationale is the same with the binomial distribution formula.

On looking back I noticed that I have mistakenly used P(4,2) in the solution. It is wrong. We've already counted different orders, so we need combination instead of permutation. Thanks for catching that.
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11 May 2005, 15:42
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( Remember this question )

Judges will select 5 finalists from the 7 contestants entered in a singing competition. The judges will then rank the contestants and award prizes to the 3 highest ranked contestants: a blue ribbon for first place, a red ribbon for second place, and a yellow ribbon for third place. How many different arrangements of prize-winners are possible?
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09 Oct 2006, 04:19
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After research I have found theoretical explanation of the combinations concept I found in the net.
Attachments

Combination.doc [118 KiB]

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13 Dec 2006, 16:07
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Rayn wrote:
How many different 6-letter arrangements ca nbe formed from the word THIRST if T must always immediately precede H?

You have T H I R S T (am not suggesting you go for a beer, itÂ´s prep time guys )

Now, T must come right before H, thus you have any of these combos:

T H X X X X

X T H X X X

X X T H X X

X X X T H X

X X X X T H

As itÂ´s not stated, you donÂ´t have to pay attention whether "1st T" or "2nd T" come before or after the other, just concentrate on having a T H paired and 4 other letters to arrange. Therefore, the answer would be

5* 4! = 5! = 120.
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26 Jan 2007, 12:19
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Quote:
Example:
Four letters are taken at random from the word AUSTRALIA. What is the probability to have two vowels and two consonants?

Solution:

Vowels: A, U, A, I, A.
Consonants: S, T, R, L

This is a case of "without replacement".
Total outcomes: C(9,4)=9*8*7*6/4!=126
Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60
Probability=60/126=10/21

I have a question about my approach. Your approach is better and fast here, however I was trying to solve this with my approach as follows and getting a different answer, I am wondering what am I missing? Can u please correct me.

P(Choosing 2 Vowels) = 5/9*4/8
P(Choosing 2 Consonants) = 4/7*3/6 (assuming no replacement)

Combine probability is = 5/9*4/8*4/7*3/6 = 5/9*1/2*4/7*1/2

Now number of ways to arrange above is 4! and hence final answer should be 4!*5/9*1/2*4/7*1/2 = 40/21

I understand that this approach is cumbersome but what is that I am doing wrong here? Why am I not getting 10/21?
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Re: Permutation, Combination and Probabilities [#permalink]

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22 Apr 2011, 11:40
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Sorry to resurrect this but is there a simpler way of answering the secretary question? (and is there an agreed answer, I've just trawled google and these boards to find several different solutions - including the 'OA' of 8/9 in the compilation available for download from this site )

Quote:
There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report?

My instinct is that the suggested answer in the file of 8/9 is far too close to 1, and starting by working out the probability that secretary A does not get a report to type <which leaves a number of other possibilities yet to remove> and subtracting from 1 gives: (2/3)^4 = 16/81 .... (which is greater than 1/9) .... 1-16/81 = 65/81

so P(everyone gets at least one) = 8/9 cannot be right.... can it? :-/

edit:: Having just put all 81 combinations into excel, there are clearly 36 ways to distribute the reports, 36/81 = 4/9) Now I just need to find the simple way of reaching that number......
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09 Mar 2005, 09:44
can you please explain differences between atleast, exactly and how to approach..
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09 Mar 2005, 09:50
The way I approach i took is.. whenever I see atleast i go ahead using 1 minus..
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09 Mar 2005, 09:57
Normally "exactly" is one case and "at least" have more than one cases. For example:
Among four babies, exactly two are boys. That means two are boys and two are girls.
Among four babies, at least two are boys. That means there maybe two boys, three boys, or four boys. The cases excluded are no boy and one boy.
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09 Mar 2005, 10:01
vprabhala wrote:
The way I approach i took is.. whenever I see atleast i go ahead using 1 minus..

Not necessarily. It depends on how convenient either approach is. For example, if asked the probability that among ten babies at least two are boys, I'd use 1 minus the probability of zero boy and one boy, but if asked the probability that among ten babies at least nine are boys, I'd simply add up the probabilities of nine boys and ten boys.
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09 Mar 2005, 10:51
makes sense..thanks guys.
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09 Mar 2005, 11:18
HongHu: That was a good tutorial ....Thanks!
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10 Mar 2005, 18:41
Thanx Hong for your Gr8 postings.

Last edited by MA on 10 Mar 2005, 20:32, edited 1 time in total.
10 Mar 2005, 18:41

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