Probability
Q: What is the probability you get an A on exam 1?
10%, 25%, 70% All arrived by personal opinion.
My estimate is 30% since in the past this has been the case. Mine is based on relative frequency
i.e. the number of students who received an A out of the number of students taking the exam.
The other probabilities are based on opinion, also called subjective probabilities. A final type of
probability is classical. This is when the there is a set of finite outcomes and the probability is
the chance a certain subset of that finite set occurs. E.g. flipping a coin and getting a Head (1 out
of 2) or rolling a die and getting a six (1 out of 6) or drawing a Heart from a deck of cards (13 out
of 52 or 1/4).
Sampling with and without replacement.
Which of the following (or both) are true?
1. Law of large numbers: Over a long run period the expected mean or proportion
of the population would be experienced. E.g. flip a coin 1000 times; we would
expect about 500 heads and tails.
2. Law of small numbers: If a pattern occurs. E.g. 6 tails in a row then we are more
likely to see a head on the next flip. Not True!
Probability is the chance that something is the case or that an event will occur. Range of
probabilities is 0 to 1.
Trial is one number of repetitions in an experiment/study.
Sample Space is the set all possible outcomes. E.g. coin flip->head, tail; die roll-> 1,2,3,4,5,6
Event is some outcome or subset of Sample Space: E.g. get a tail on a coin flip; E.g. get an
even number when die rolled.
Complement is a set of outcomes not in the event of interest. E.g. if event of interest is “get a tail
on coin flip” a complement is “not a tail” I.E. get a head. E.g. If event is get an even number on a
roll of die” I.E. we get a 2 or 4 or 6. Complement would be 1, 3, or 5. If the event is “pass the
class with C or better” then the complement is not pass the class meaning get a D or F….NOTE:
getting just a D or getting just an F are not the complement but only part of the complement. The
complement consists of ALL outcomes NOT in the event of interest.
Intersection is when two or more events overlap. i.e. share common outcomes. E.g. let event E
be get even number on die roll (2,4,6), Let S be event we get a 6.
Union: the event of all outcomes of two or more events. E.g. Let E be even numbers on role of
die, let F be get a 5, the union of events E and F would be (2,4,5,6)
Conditional Probability is the probability that event occurs given that another event has already
happened. E.g. Let E be event get an even number on roll of die. Say we know the outcome is an
even number, what is probability we got a 6? 1/3
Independent trials one event does not influence the other event. E.g. coin flip.
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Mutually Exclusive of Disjoint Events: events that do not have any common outcomes. Both
events cannot occur at the same time. E.g. let E be even number on roll of die, Let F be we get a
5.
Simple Events: are events of only one outcome. E.g. getting a 5.
Notation:
If we define two events as A, and B then
P(A): is probability that event A occurs
P(AC ) or P(A1): the complement of A
P(AorB) or P(AUB: The union of A and B]
A stat class has 4 teaching assistants.
3 females( Lauren, Rona, Leila)
1 male (Tom)
Each Ta is on chare of one section, Let event W be the Ta is female and let event M be TA is
male.
1) Which event is a simple event? Event M
2) Are the events W and M mutually exclusive? Yes Since the events do not overlap. I.E.
share any common outcomes.
3) If two students, unknown to each other, randomly select a TA are their choices
independent? Yes
4) What is probability that a student randomly selects Lauren? ¼
5) You are told your section TA is female. What is probability that Lauren is your TA?
1/3
Mathematical Rules:
- 0 <= P(event) <= 1
- P(event) + P(event complement) = 1
- P(A U B) = P(A) + P(B) - P(A and B)
- If two events, say A and B, are independent then P(A and B) = P(A)*P(B)
- Conditional Probability: P(A|B) means “probability event A occurs given event B has
already occurred.” P(A|B) = P(A and B)/P(B)
- Alternatively, P(B|A) = P(A and B)/P(A)
Note then that if the two events are independent then P(A|B) = P(A) and P(B|A) = P(B) For
example, what is the probability that on two draws from a card deck that both cards are Aces?
That is, find probability that the first card is Ace AND second card is an Ace? Well probability
of first card is Ace is 4/52 or 1/13. Then for both to be an ace we can find the conditional
probability of second card is Ace given first card was and Ace. This conditional probability is
3/51. So P(first and second are Aces) equals the 1/13*3/51 or 3/663
Terminology
Probability: The likelihood that some outcome occurs. The range of any probability is from 0 to
1. The most common calculation of this value comes from the proportion of times that the
outcome would occur in a long run of observations.
Trial: One of a number of repetitions of an experiment.
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Independent Trials: Trials are considered independent if the outcome of any one trial is not
affected/influenced by the outcome of any other trial.
Sample Space: The set of all possible outcomes. The sum of all individual probabilities within a
sample space is one.
Event: A subset of the sample space.
Complement: The complement of an event consists of all outcomes in the sample space that are
not in that event.
Intersection: The intersection of two events consists of all outcomes that in both events.
Pictorially, this means that the two events overlap.
Union: The union of two events consists of all outcomes that are either in the one event and/or in
the other event.
Conditional Probability: Conditional probability is the probability that an event occurs given
that another event has already occurred.
Mutually Exclusive or Disjoint: Two events are considered mutually exclusive, or disjoint, if
they do not share any common outcomes. Pictorially, this means that the two events do not
overlap.
Notation
If we define two events as A and B then;
P(A): is the probability that event A occurs.
P(Ac): is the complement to A
P(A or B) of P(A U B): The union of events A and B
P(A and B) of P(A ∩ B): The intersection of events A and B
P(A|B): The conditional probability of event A occurring given that event B has occurred. The
vertical slash “|” represents given. The formula for finding conditional probability is:
P(A|B)= P(A and B)/P(B) [read the “probability of A and B, divided by probability of B.”]
Independence and Union of Two Events
Two events, say event A and event B, are independent if any one of the following can be proven
true.
P(A and B) = P(A)*P(B)
P(A|B) = P(A)
P(B|A) = P(B)
The union of these two events is found as follows:
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P(A or B) = P(A) + P(B) – P(A and B). [note that if A and B are mutually exclusive the P(A and
B) is zero.
Scenarios
Illustrate how common sense applies:
Example 1: With a die ask what is probability of getting a 6? Answer 1/6 and demonstrate this by
letting event A = “getting a 6” and P(A) = 1/6
Example 2: With the die what is probability of getting an even number? Answer 3/6 or 1/2 where
we let event B = “getting and even number” and P(B) = 1/2
Example 3: Say I roll the die and tell you that the number is even (i.e. given you this event
occurred ---- conditioned on this information) what is the probability that the number is a 6?
Answer, with the given information of being even we are down to 3 possible outcomes: 2, 4, or 6
and from this only one outcome is the event of interest a 6 making the probability of getting a 6
when knowing the number is even is 1/3. Using the notation and formula, we let event A =
getting a 6 and let B = getting an even number. From our conditional formula we have P(A|B) =
P(A and B)/P(B) = (1/6)/1/2) = 2/6 = 1/3
Pre-Class Probability Examples
1. You apply to two graduate schools: Harvard and Pitt. The probability that you get accepted
into Harvard is 0.3, the probability you get into Pitt is 0.7, and the probability you get accepted
into both Harvard and Pitt is 0.25 Are the two individual events – getting accepted into Harvard
and getting accepted into Pitt – considered independent events? Why or why not?
Dependent because P(H)*P(P) = P(H and P) that is: 0.3*0.7 = 0.21 which does not equal 0.25
2. You have a six-sided die consisting of green on four sides and red on the other two. You roll
the die five times. Which of the following sequences is MOST likely to happen?
- Green, Green, Green, Green, Green
- Green, Red, Green, Red, Green
- Green, Green, Green, Red, Green
The sequence with the most green since green is most likely to appear after each toss.
3. The football game is almost over. Your team has the ball and is winning by four points
meaning the other team needs a touchdown to beat you. It is fourth down and you need two yards
for a first down. You have two choices:
1. Go for the first down. If you are successful you win. The chances of getting the first
down are 60%. If the play fails, then the other team gets the ball with a 53% chance of scoring
the winning touchdown.
2. Or you can kick the ball to the other team. By choosing this option your team has a
70% chance of winning.
Which choice should you make? That is, which of the two options results in the greater
probability that your team wins?
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Probability win by going for if kick is 0.70 so you would not kick if probability of winning
for this choice exceeds 0.70
Probability of win if no kick is found by:
(0.6)*(1) + (0.4)(1-0.53) = 0.6 + 0.188 = 0.788
Since probability of winning for not kicking is greater than probability of winning if kicked,
the decision would be to not kick.
4. Back in the 1970s there was a game show called "Let's Make a Deal". The premise of the
game was this: A contestant was shown three curtains behind one of which was a real prize (e.g.
money, new car) while the other two hid gag prizes (e.g. donkey eating hay). The rules were
quite simple. The contestant picked a door (e.g. curtain #1) and the host then opened one of the
other curtains behind which was a gag gift (e.g. curtain 3). This left the contestant with two
curtains still remaining; the one chosen and the other. The host then offered the contestant the
option to either stick with their original curtain or switch to the other. Question: If you were the
contestant, would you switch at this stage of the game or stay with your first choice? That is,
which decision - staying with first choice or switching to other curtain – gives you a better chance
of winning?
You Pick
1
1
1
2
2
2
3
3
3
Prize Door
1
2
3
1
2
3
1
2
3
Monty Opens
2 or 3
3
2
3
1 or 3
1
2
1
1 or 2
Don’t Switch
Win
Lose
Lose
Lose
Win
Lose
Lose
Lose
Win
3 Wins (33%)
Switch
Lose
Win
Win
Win
Lose
Win
Win
Win
Lose
6 Wins (66%)
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