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VP  Joined: 03 Jan 2005
Posts: 1394
I'm going to try to collect some P and C type questions and  [#permalink]

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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.

Originally posted by HongHu on 09 Mar 2005, 07:41.
Last edited by HongHu on 09 Mar 2005, 08:33, edited 1 time in total.
VP  Joined: 03 Jan 2005
Posts: 1394

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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
##### General Discussion
VP  Joined: 03 Jan 2005
Posts: 1394

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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?
Director  Joined: 30 Sep 2004
Posts: 997
Location: Germany

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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
VP  Joined: 03 Jan 2005
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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.

Originally posted by HongHu on 09 Mar 2005, 14:57.
Last edited by HongHu on 25 Apr 2011, 04:50, edited 2 times in total.
Intern  Joined: 06 Feb 2006
Posts: 12
I disagree with honghu solution  [#permalink]

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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

This is my thinking how the solution should be:
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:
Chose 1 sectretary and chose 2 reports = C(3,1)*C(4*2)
then, chose 1 secretary(out of two) and 1 report(out of two) = C(2,1)*C(2,1)
then, chose 1 secretary(out of one) and 1 report(out of one) = C(1,1)*C(1,1)

net combinations where each secretary has atleast 1 report is multiple of all three above = C(3,1)*C(4*2)*C(2,1)*C(2,1)*C(1,1)*C(1,1)
3*6*2*2*1*1 = 72
So the probability is 72/81.

I would appreciate if HongHu or someone else can point out if there is any error in my solution and why.
Intern  Joined: 08 Oct 2006
Posts: 23

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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]

Intern  Joined: 08 Oct 2006
Posts: 23

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permutations theoretical concept (sorry first time the file hasn't attached)
Attachments Permutation.doc [78.5 KiB]

Non-Human User Joined: 09 Sep 2013
Posts: 15418
Re: I'm going to try to collect some P and C type questions and  [#permalink]

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_________________ Re: I'm going to try to collect some P and C type questions and   [#permalink] 30 Jun 2019, 17:58

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