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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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Binomial Distribution [#permalink]
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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 1p. What is the probability of A happening exactly k times in n repeated tests?
Formula:
C(n,k) * p^k * (1p) ^ nk
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=1C(10,0)*0.5^0*0.5^10C(10,1)*0.5^1*0.5^9
=10.5^1010*0.5^10
=111*0.5^10



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Repeated Outcomes [#permalink]
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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 = 2^91
(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 = (2^31)*(2^41)*(2^21)
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?



Director
Joined: 21 Sep 2004
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can you please explain differences between atleast, exactly and how to approach..



Director
Joined: 21 Sep 2004
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The way I approach i took is.. whenever I see atleast i go ahead using 1 minus..



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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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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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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 = 2^91
(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



Director
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makes sense..thanks guys.



Director
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HongHu: That was a good tutorial ....Thanks!



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Letters and Words [#permalink]
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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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Thanx Hong for your Gr8 postings.
Last edited by MA on 10 Mar 2005, 20:32, edited 1 time in total.



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question? [#permalink]
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10 Mar 2005, 20:17
For Outcomes with two vowels and two consonants, why isn't it: "4!" instead of P(4,2)?
================================================
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: P(4,2)*5^2*4^2
You can calculate the probability from here.



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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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Question: in the above example, why isn't total outcome = 4^3?
Hong in the 3 secretaries and 4 departments problem why di we get the solution as 3^4 and not 4^3? Please explain the concept
Thanks



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gmat2me2 wrote: Question: in the above example, why isn't total outcome = 4^3?
I'm not sure if you are talking about the AUSTRALIA problem. In that problem we need a word that have 4 letters, and for each letter there are 9 possible choices, so the total outcome is 9*9*9*9=9^4. Quote: Hong in the 3 secretaries and 4 departments problem why di we get the solution as 3^4 and not 4^3? Please explain the concept
This is a very good question. In this question each secretary can have 0 to 4 reports to type, but each report must be and can only be typed by one person. Here you need to look which one is replaced back when you repeat the test. If a report is typed by one person, then it will not be typed by another person. In other words, the reports can not be replaced. On the other hand, if a secretary has typed one report, she'll still be placed in the pool for the second report. In other words, the secretary is the one that gets replaced when we repeat the test, and each report faces the same pool of three secretaries. Therefore, the outcome is 3*3*3*3 for the four reports.
Let me give you another example, see if it will help.
Example:
1. Five rooms are to be allocated to four people. How many ways are there if each room must be assigned to one person and one person only?
Here each people can have more than one room, or have no room at all. In other words the people will be replaced back into the pool in each repeated test and each room will face the same pool of people. So five room each face a pool of four people. 4*4*4*4*4=4^5
2. Five rooms are to be allocated to four people. How many ways are there if each person must be assigned to one room and one room only?
Here each room can be empty, or can have multiple people assigned to it. In other words the rooms will be placed back into the pool for each repeated test and each person will face same pool of five rooms. Therefore total outcome is 5*5*5*5=5^4.



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Re: Binomial Distribution [#permalink]
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26 Mar 2005, 13:37
HongHu wrote: Binomial Distribution
In each individual test, the probability of A happening is p and not happening is 1p. What is the probability of A happening exactly k times in n repeated tests?
Formula: C(n,k) * p^k * (1p) ^ nk
How can I count C(n,k)?
Thank you



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Re: Binomial Distribution [#permalink]
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31 Mar 2005, 23:49
billy bonse wrote: HongHu wrote: Binomial Distribution
In each individual test, the probability of A happening is p and not happening is 1p. What is the probability of A happening exactly k times in n repeated tests?
Formula: C(n,k) * p^k * (1p) ^ nk
How can I count C(n,k)? Thank you
C(n,k) = n!/k!(nk)!



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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 prizewinners are possible?
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I thought I have this here but I guess I haven't ...
The no. of ways to distribute n items among r people where each gets zero or more items is (n+r1) C(r1).
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