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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.
09 Mar 2005, 08:54
Kudos
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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
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
09 Mar 2005, 09:29
Kudos
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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?
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?
