Question: What is the difference between counting and finding probability?
When measuring discrete (whole number) outcomes, the classical definition of a probability of a desired outcome \(A\) is:
\[P(A) = \frac{ \text{number of outcomes in which A occurs}}{\text{total number of possible outcomes}}\]
Quiz Question: If you roll one die, what is the probability of getting a 6?
Quiz Question: If you roll two dice, what is the probability of getting a sum of 6?
Sometimes probabilities are easier to calculate if we look at their complement.
The complement of an event \(A\) is the event “\(A\) doesn’t happen.” The notation \(A^c\) is used for the complement of event \(A\). We can compute the probability of the complement using: \[P(A^c) = 1 - P(A)\]
Note: The complement of \(A^c\) is the original event \(A\), so that \[P(A) = 1 - P(A^c)\]
Quiz Question: If you roll one die, what is the probability of not getting a 6?
Quiz Question: If you roll two dice, what is the probability of not getting a sum of 6?
Example 1 - How many different ways are there to line up 5 students?
5*4*3*2*1## [1] 120Factorial:
\[n! = n(n-1)(n-2) ... 3 \cdot 2 \cdot 1\]
n = 5
factorial(n)## [1] 120Quiz Question:
How many different ways can you arrange red, blue, green, and yellow
balls in a line?
Example: - How many ways different ways can I pick 3 students from a class of 20 and put them in a row?
20*19*18## [1] 6840Permutations: The number of ways \(r\) items may be selected from among \(n\) choices (without replacement) when order matters is:
\[_n P_r = n(n-1)(n-2) ... (n-r+1)\] \[ = \frac{n(n-1)(n-2) ... (n-r+1)}{1} \cdot \frac{(n-r)(n-r-1) ... 3 \cdot 2 \cdot 1}{(n-r)(n-r-1) ... 3 \cdot 2 \cdot 1} \] \[ = \frac{n!}{(n-r)!}\]
NOTE: Many calculators have the function ‘nPr’ for the number of permutations. Thus, an easier way to calculate \(\frac{n!}{(n-r)!}\) is simply: \(nPr(n,r)\).
If you use R, here is some code that makes your own nPr function:
# Fast permute function
nPr = function(n,k) {
sum = n
for ( i in 1:(k-1) ) {
sum = sum*(n-i)
}
sum
}
# Example computation
nPr(20,3)## [1] 6840Quiz Question: How many ways can a four-person executive committee (president, vice-president, secretary, treasurer) be selected from a 16-member board of directors of a non-profit organization?
Example 3 - How many ways can a four-person committee be selected from 16 members where all committee members have equal positions?
There are \(_{16} P_{4}=43680\) ways to choose the members where the order matters. If it doesn’t, we overcounted. By how much? For any 4 member selection, there are \(4!\) ways to order those 4 members, thus we overcounted by a factor of \(4!\)
# Example 3
43680/factorial(4)## [1] 1820Combinations: The number of ways \(r\) items may be selected from among \(n\) choices (without replacement) when order does NOT matter is: \[_n C_r = \frac{_n P_r}{r!} = \frac{n!}{r!(n-r)!}\] This is also denoted \(\binom{n}{r}\) and referred to as “n choose r.”
So we could be fancy and calculate it like this:
# Example 3
n = 16
r = 4
factorial(n)/( factorial(r) * factorial(n-r) )## [1] 1820Or even this!
# Example 3
choose(16,4)## [1] 1820NOTE: Many calculators have the function ‘nCr’ for the number of combinations Thus, an easier way to calculate \(\frac{n!}{(n-r)!r!}\) is simply: \(nCr(n,r)\).
Quiz Question:
A group of four students is to be chosen from a 35-member class to
represent the class on the student council. How many ways can this be
done?
Consider the likelihood of ANY two people from a class of 60 students having the same birthday.
To find the chance of ANY two people in this class having the same birthday, let’s break this up into easier problems as follows.
Question: What is the probability that, given just two people selected, they have the same birthdays?
If there are \(365\) different birthdays:
\[P(\text{two given people have the same birthday}) = \frac{365}{365^2} = \frac{1}{365}\]
1/365## [1] 0.002739726Let \(A\) = “no 2 people out of 3 have the same birthday”: Then \(A^c\) = “at least 2 people out of 3 share the same birthday, so:
\(P(\text{at least 2 people out of the 3 have the same birthday})\) \[= 1-P(\text{no 2 people out of 3 have the same birthday})\]
\(P(\text{no 2 people share the same birthday})\) \[=\frac{\text{number of different ways 3 people could have all different birthdays}}{\text{total ways 3 people could have birthdays}}\]
The probability that no 2 out of 3 people share the same birthday:
365*364*363/365^3## [1] 0.9917958What is the probability that, given 3 people, at least one pair of people share the same birthday?
What is the probability that, given 5 people, at least one pair of people the same birthday?
Hint: \(P(\text{no 2 people out of 5 share same birthday})=\frac{_{365} P_{5} }{365^{5}}\)
Hint: \(P(\text{no 2 people out of 60 share same birthday}) = \frac{_{365} P_{60} }{365^{60}}\)
Two events are independent if the outcome of one does not affect the probability of the other. If events A and B are independent, then the probability of both \(A\) and \(B\) occurring is \[P(A \text{ and } B) = P(A) \cdot P(B)\] where \(P(A \text{ and } B)\) is the probability of events \(A\) and \(B\) both occurring, \(P(A)\) is the probability of event A occurring, and \(P(B)\) is the probability of event \(B\) occurring.
If A is the event that the first machine is broken and B is the event that the second machine is broken, the probability both are broken is: \[P(A \text{ and } B) = P(A) \cdot P(B)\] (If we assume A and B are independent events.)
# a. Find the probability that at least one bottling system is working (reliability) is the complement of this answer.
# b. Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin or a 6 on the die.
Here, there are still 12 possible outcomes: {H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}
How many outcomes have heads?
How many outcomes have a 6?
How many outcomes have heads OR a 6?
How many outcomes have heads AND a 6?
The probability of either of two events occurring simultaneously is \[P(A \text{ or } B)=P(A)+P(B)-P(A \text{ and } B)\].
Question: Suppose you draw one card from a standard card deck. What is the probability of getting a club or a face card?
Quiz Question: Suppose you draw one card from a standard card deck. What is the probability of getting a spade or an ace?
Two events are mutually exclusive if they cannot happen at the same time, so \(P(A \text{ and } B) = 0\). If A and B are mutually exclusive, then \[P(A \text{ or } B) = P(A) + P(B)\]
Question: If you roll a 6-sided die, what is the probability of rolling an even number?
Suppose we draw one card from a standard card deck. What is the probability that we get a Queen or a King?
Suppose we draw one card from a standard card deck. What is the probability that we get a spade or a King?
A multiple-choice question on a quiz contains 5 questions, each with four possible answers (A, B, C, D). Compute the probability of randomly guessing the answers and getting 100% on the quiz (all five questions correct).
Question 1:
What is the total number of possible ways you could respond to this
test?
Question 2:
What is the probability of getting all correct?
Question 3:
What is the probability of guessing and getting 0 questions correct (all
incorrect)?
Question 4.1:
What is the probability of guessing and getting 1 question correct?
Question 4.2:
What is the probability of guessing and getting 2 questions correct?
Question 4.3:
What is the probability of guessing and getting 3 questions correct?
Question 4.4
What is the probability of guessing and getting 4 questions correct?
a = c()
a[1] = 3^5/4^5
a[2] = choose(5,1)*3^4/4^5
a[3] = choose(5,2)*3^3/4^5
a[4] = choose(5,3)*3^2/4^5
a[5] = choose(5,4)*3^1/4^5
a[6] = 1/4^5
names(a) = paste( "P(", c(0:5), " correct)", sep="")
a## P(0 correct) P(1 correct) P(2 correct) P(3 correct) P(4 correct) P(5 correct)
## 0.2373046875 0.3955078125 0.2636718750 0.0878906250 0.0146484375 0.0009765625sum(a)## [1] 1Question:
On the quiz, what is the probability of getting a “D” or higher (at
least 3 out of 5 answers correct)?
P(getting 3, 4, or 5 correct) = P(3 correct) + P(4 correct) + P(5 correct})
Expected value provides a way of evaluating the value of a decision with multiple outcomes.
Expected Value is defined as the average gain or loss of an event if the procedure is repeated many times. We can compute the expected value by multiplying each outcome by the probability of that outcome, and then adding up the products.
For mutually exclusive events, A and B, the expected value is:
\(P(A)\cdot V(A) + P(B)\cdot V(B)\),
where \(V(A)\) and \(V(B)\) represent the value of \(A\) and \(B\) respectively, with a gain as a positive value and loss as a negative value.
For \(n\) disjoint events \(A_1, A_2, ... A_n\) for which \(P(A_1) + P(A_2) + ... + P(A_n)=1\), the expected value is:
\(P(A_1)\cdot V(A_1) + P(A_2) \cdot V(A_2) + ... + P(A_n)\cdot V(A_n)\).
Example: You purchase a raffle ticket to help out a charity. The raffle ticket costs $5. The charity is selling 2000 tickets. One of them will be drawn and the person holding the ticket will be given a prize worth $4000. Compute the expected value for this raffle.
So the expected value is:
# Example 8
3995*1/2000 + -5*1999/2000## [1] -3On average, each person is giving about $3.00 to charity.
An insurance company estimates the probability of an earthquake in the next year to be 0.0013. The average damage done by an earthquake is estimated to be $60,000. If the company offers earthquake insurance for $100, what is the expected value of the policy?
A friend offers to play a game, in which you roll 3 standard 6-sided dice. If all the dice roll different values, you give him $1. If any two dice match values, you get $2. What is the expected value of this game? Would you play?
P_lose = 6*5*4/6^3
(P_lose)*(-1) + (1-P_lose)*(2)## [1] 0.3333333The probability that event B occurs, given that event A has happened is represented by \(P(B|A)\), read “the probability of B given A.”
Conditional probabilities can be used to find the probability of joint events, even when they are not independent:
\[P(A \text{ and } B) = P(A|B) \cdot P(B)\]
Consider the following events:
## Class No Class Sum
## Power 90 6 96
## No Power 0 4 4
## Sum 90 10 100What is the probability of having class?
Question:
What is the probability of not having class?
Question:
What is the probability of having class given that
there was no power when you wake up?
Question:
What is the probability of having class given that
there was power when you wake up?
Quiz Question: Given that there was class, what is the probability that there was power?
Quiz Question:
Given that there was no class, what is the probability there was no
power?
By the the formula for calculating join probabilities (for events that are not dependent) both \[P(A \text{ and } B) = P(A|B) \cdot P(B)\] and \[P(B \text{ and } A) = P(B|A) \cdot P(A)\] By setting these equal we get a way to “invert” conditional probabilities: \[P(A|B) \cdot P(B)=P(B|A) \cdot P(A)\] OR
\[P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\]
If we only know \(P(A)\) and \(P(B|A)\), we can find \(P(B)\) because:
This accounts of all the ways \(B\) can occur, so \[P(B) = P(B|A) \cdot P(A) + P(B|A^c) \cdot P(A^c)\].
Now, plugging in \(P(B)\) gives us Bayes’ Rule!
Bayes’ Rule: Given two events \(A\) and \(B\),
\[ P(A|B) = \frac{P(B|A) \cdot P(A)} { P(B|A) \cdot P(A) + P(B|A^c) \cdot P(A^c)} \]
Example 9 - A new test has been devised to detect a new disease:
Should doctors use this test?
Well, suppose you test negative for the disease. Great, that means you don’t have it! But if you test positive, what is the probability that you actually have the disease?
Let’s first label the following events:
So we want to know \(P(\)having the disease|testing positive\() =P(A|B)\).
What information do we know?
-\(P(\)having the disease\()=P(A)=\)
-\(P(\)not having disease\(=P(A^c)=\)
-\(P(\)testing negative|have the disease\()=P(B^c|A)=\)
-\(P(\)testing positive|have the disease\()=P(B|A)=\)
-\(P(\)testing positive|not having the disease\()=P(B|A^c)=\)
-\(P(\)testing negative|not having the disease\()=P(B^c|A^c)=\)
Using Bayes’ Rule:
\[ P(A|B) = \frac{P(B|A) \cdot P(A)} { P(B|A) \cdot P(A) + P(B|A^c) \cdot P(A^c)} \]
# Example 9
(1*0.001)/(1*0.001+0.05*0.999)## [1] 0.01962709This shows that only about 2% of the people who test positive for this disease using this test will actually have it!
A new test has been devised to detect a new disease:
Again, label the following events:
If you test positive, what is the probability of actually having the disease?
If you test negative, what is the probability of not having the disease?
Bayes’ rule can be used for this if we replace \(A\) with \(A^c\), replace \(B\) with \(B^c\), and realize that the complement of \(A^c\) is just \(A\).
\[ P(A^c|B^c) = \frac{P(B^c|A^c) \cdot P(A^c)} { P(B^c|A^c) \cdot P(A^c) + P(B^c|A) \cdot P(A)} \]
In your subgroup, select a problem you want to work on. Work together toward finding a solution to the provided questions and/or any related questions you find interesting. In a 1-2 page Report, present:
Make sure you edit your Report to ensure it is readable with no grammar or spelling errors.
Each group will also make a short video presentation of their work, so keep in mind, that your work will be made public for other students to view and study.
Compute the probability of randomly drawing five cards from a deck and getting:
a. a pair
b. three-of-a-kind
c. four of a kind
d. a full house (three of a kind and a pair)
e. a flush (all the same suit)After you have answers your group is convinced of your answers, try checking your answers: https://en.wikipedia.org/wiki/Poker_probability
Suppose you have three of a kind. What is the probability that someone else has a higher hand? (you can use the probabilities given for a straight and straight flush on Wikipedia for answering this one.)
In a certain state’s lottery, \(64\) balls numbered 1 through \(64\) are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins a jackpot of $1,000,000. If numbers drawn match any 5 of the numbers that a player had chosen, the player wins $1,000. It costs $1 to buy a ticket. Find the expected value.
Over time the jackpot will increase if no one wins. How large would the jackpot have to be for the expected value of playing the lottery to be positive? (In this case, would you still buy a lottery ticket?)
Suppose you had problems working with your group on last projects unit. Some were arguing and some were not contributing to the work on the group, so was decided that students will be randomly placed into 6 groups on the next project. Should you worry about being put into the same group??
Use the following questions to develop an answer to the question: If we were put to randomly groups again, what is the probability that YOU are placed in the exact same group?
How many different ways are there to place \(n=30\) students into 6 groups? (Hint: consider an easier problem.)
If we were to randomly assign groups again, what is the probability EVERYONE is placed in the exact same groups again?
If you are put in the same group, how many ways are there to assign everybody else? Use this to find the probability that, if we are randomly assigned into new groups, YOU are placed in the exact same group.
In this problem, we will explore probabilities from a series of events.
If you flip 10 coins, how many would you expect to come up “heads” on average? Given your answer, would you expect every flip of 10 coins to come up with exactly that many heads?
If you were to flip 10 coins, what results would you consider “usual”? What results would you consider “unusual”?
When flipping 10 coins, what is the theoretic probability of flipping 10 heads?
Below is a simulation of flipping a coin 10 times, repeated 100,000 times. In the table below, the number of times 0,1,2,…,10 heads were displayed. Note the sum of all these values is 100,000.
outcomes = c("heads","tails")
trial = c()
for ( i in 1:100000 )
{
sam = sample( outcomes, replace = T, 10)
trial[i] = sum( sam=="heads" )
}
kable(table(trial), caption="Table of number of time out of 100,000 flips",
col.names = c("Number of heads", "Number of trials"))| Number of heads | Number of trials |
|---|---|
| 0 | 89 |
| 1 | 1001 |
| 2 | 4414 |
| 3 | 11756 |
| 4 | 20508 |
| 5 | 24751 |
| 6 | 20431 |
| 7 | 11544 |
| 8 | 4432 |
| 9 | 992 |
| 10 | 82 |
Based on this simulation, what appears to be the probability of flipping 0 heads, 1 head, … up to 10 heads?
If you were to flip 10 coins, based on the simulated data, what range of values would you consider “usual” results? What would you consider “unusual” results?
The formula \[_n C_k p^k (1-p)^{n-k}\] will compute the probability of an event with probability \(p\) occurring \(k\) times out of \(n\), such as flipping \(k=5\) heads out of \(n=10\) coins where the probability of heads is \(p=0.5\).
\[_n C_r = \frac{_n P_k}{k!} = \frac{n!}{k!(n-k)!}\] You can use the Excel formula “=COMBIN(n,k)” to calculate \(_nC_R\).
Use this to compute the theoretic probability of flipping 5 “heads” out of 10 coins. Compare your answer to the probability found in 4.
Use this to compute the theoretic probability of flipping fewer than 2 “heads” out of 10 coins. Compare your answer to the probability found in 4.
Use this formula to consider a case from 1960. In the area, about 26% of the jury-eligible population was black. In the court case, there were 25 people on the juror panel, of which 2 were black.
If black people were selected for the panel with a probability of 26%, calculate the probability of there being 2 or fewer black people on the jury panel.
Does this provide evidence of racial bias in jury selection?
An unmanned monitoring system uses high-tech video equipment and microprocessors to detect intruders. A prototype system has been developed and is in use outdoors at a weapons munitions plant. The system is designed to detect intruders with a probability of .90. However, the design engineers expect this probability to vary with the weather conditions. The system automatically records the weather conditions each time an intruder is detected. Based on a series of controlled tests, in which an intruder was released at the plant under various weather conditions, the following information is available: Given the intruder was, in fact, detected by the system, the weather was clear 75% of the time, cloudy 20% of the time, and raining 5% of the time. When the system failed to detect the intruder, 60% of the days were clear, 30% cloudy, and 10% rainy. Use this information to find the probability of detecting an intruder, given clear, cloudy, and rainy weather conditions (in the case that an intruder has been released at the plant). When is this system the most reliable? When is it the least reliable?
Suppose a certain type of cancer has an incidence rate of 0.5% (that is, it afflicts 0.5% of the population). A new test has been devised to detect this cancer, which is very cheap and easy to administer in comparison to existing tests. The test produces false negatives at a rate of 1.4% (that is, 1.4% of those who have the disease will test negative), and the false positive rate is 1% (that is, about 1% of people who take the test will test positive, even though they do not have the disease). How accurate is this test?
Based on this, what recommendations would you make to doctors using this test?