Stat 250 Gunderson Lecture Notes
Chapter 7: Probability
Chance favors prepared minds. -- Louis Pasteur
Many decisions that we make involve uncertainty and the evaluation of probabilities. Example: Roll a fair die possible outcomes = { 1, 2, 3, 4, 5, 6 }
Before you roll the die do you know which one will occur? No
What is the probability that the outcome will be a ‘4’? 1/6 = P(‘4’)
Why?
A few ways to think about PROBABILITY:
(1) Personal or Subjective Probability
P(A) = the degree to which a given individual believes that the event A will happen.
(2) Long term relative frequency
P(A) = proportion of times ‘A’ occurs if the random experiment (circumstance) is repeated
many, many times.
P(A) = proportion of balls
in the basket that have
an ‘A’ on them.
Note: each time I do the experiment,
Note: A probability statement IS NOT a statement about INDIVIDUALS .
It IS a statement about the population / the basket of balls .
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7.3 Probability Definitions and Relationships
7.4 Basic Rules for Finding Probability
There is a lot you can learn about probability. One basic rule to always keep in mind is that the probability of any outcome is always between 0 and 1. Now, there are entire cours devoted just to studying probability. But this is a Statistics class. So rather than start with a list of definitions and formulas for finding probabilities, let’s just do it through an example so you can e what ideas about probability we need to know for doing statistics.
Example: Shopping Online
Many Internet urs shop online. Consider a population of 1000 customers that shopped online at a particular website during the past holiday ason and their results regarding whether or not they were satisfied with the experience and whether or not they received the products on time. The results are summarized below in table form. Using the idea of probability as a proportion, try
a. What is the probability that a randomly lected customer was satisfied with the experience?
Think about probability as proportion, so 820/1000 = 0.82 or 82%
This would be reprented by P(Satisified).
b. What is the probability that a randomly lected customer was not satisfied with the experience?
Well if 82% were satisfied, then 1 – 0.82 = 0.18 or 18% were not satisfied. Would be reprented by P(Not Satisfied), and you just ud what we call the complement rule.
c. What is the probability that a randomly lected customer was both satisfied and received the
product on time?
Again find the proportion of customers that have both of the traits
… 800/1000 = 0.80 or 80%. This would be reprented by P(Sat and OT).
d. What is the probability that a randomly lected customer was either satisfied or received the
product on time? Here we have an “OR”. So we need the proportion of customers that were either satisfied or received it on time. Start with how many were satisfied (820) then let’s add in tho that received it on time (880), but in doing so, we included tho that were both satisfied and received on time (800) twice! So we need to correct for that double counting…
(820 + 880 – 800) / 100 = 900/1000 = 0.90 or 90%.
P(satisfied or on time)=P(satisfied)+P(on time) – P(satisfied and on time) = 0.82+0.88–0.80=0.90 e. Given that a customer did receive the product on time, what is the probability that the customer
was satisfied with the experience?
You can just u the table to logically figure it out. We should only look at tho customers that received it on time … there were 880. Out of the 880, let’s find the probability (the
proportion) they were satisfied… 800 out of 880 or 0.901 (90.1%)
What you just found was a conditional probability: P(Satisfied|on time) where the line means “Given” or “Conditional on” or “among tho”
f. Given that a customer did not receive the product on time, what is the probability that the
customer was satisfied with the experience?
Out of the 120 that did not receive on time, the proportion that were satisfied… 20 out of 120 or 0.167 (16.7%). This conditional probability is P(Satisfied|Not on time); note how much lower it is compared to part (a) and part (e).
Note: We stated that the above 1000 customers reprented a population. If results were bad on a sample that is reprentative of a larger population, then the obrved sample proportions would be ud as approximate probabilities for a randomly lected person from the larger population.
Great job! You just computed probabilities using many of the basic probability rules or formulas
You did not need the formulas themlves but instead ud intuition and approaching it as some type of proportion. Let’s e how your intuition and the above formulas really do connect.
In part b you found the probability of “NOT being satisfied”, which is the complement of the event “being satisfied”, so the answer to part b is the complement of the probability you found in part a.
In part c, there was a key word of “AND” in the question being
asked. The “AND” is just the interction, or the overlapping
part; the outcomes that are in common. The picture at the right
show an interction between the event A and B. In a table, the
counts that are in the middle are the “AND” counts; there were 800 (out of the 1000 customers) that were both a multiplication formula above for finding probabilities of the AND or interction of two events, but we did not even need to apply it; as a table prentation of counts provides “AND” counts directly.
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In part d, there was a key word of “OR” in the question being asked. The “OR” is union, the outcomes that are in either one or the other (including tho that are in
both). The picture at the right show an union between the event
A and B. Notice that if you start with all of the outcomes that are
in A and then add all of the outcomes that are in B, you have
double counted the outcomes that are in the overlap. So the
addition formula above shows you need to subtract off the
intercting probability once to correct for the double counting. From the table, you could either add up the parate counts of 800 + 20 + 80; or start with the 820 that were satisfied and add the 880 that received it on time and then subtract the 800 that were in both ts; to get the 900 in all that wer
e either satisfied or received the product on time.
Finally, parts e and f were both conditional probabilities. In part e you were first told to consider only the 880 customers that received the product on time, and out of the find the probability (or proportion) that were satisfied. There were 800 out of the 880 that were satisfied. The picture below shows the idea of a conditional probability formula above for P(A | B), read as the probability of A given B has occurred. If we know B has occurred, then only look at tho items in the event B. The event B, shaded at the right, is our new ‘ba’ (and thus is in
the denominator of the formula). Now out of tho items in B,
we want to find the probability of A. The only items in A that are
手机指纹on the t B are tho in the overlap or interction. So the
conditional formula above shows you count up tho in the “A and B” and divide it by the ba of “B”.
管仲破厚葬Try It! Go back to parts a to f and add the corresponding shorthand probability notation of what you actually found; e.g. P(satisfied), P(satisfied | on time) next to each answer.
Now there are a couple of uful situations that can make computing probabilities easier.
We can easily picture disjoint events becau the definition is a property about the ts themlves.
If A, B are disjoint, then P(A and B) = 0. If there are no items in the overlapping part, then many of th
e probability results will simplify. For example, the additional rule for disjoint events is P(A or B) = P(A) + P(B).小学家访记录
Another important situation in statistics occurs when the two events turn out to be independent.
降血压的最好办法This expression P(A|B) = P(A) tells us that knowing the event B occurred does not change the probability of the event A happening. Now it works the other way around too, namely, if A and B are independent events, then P(B|A) = P(B).
As a result of this independence definition, we could show that the multiplication rule for independent events reduces to P(A and B) = P(A)P(B). Finally, this rule can also be extended. If three events A, B, C are all independent then P(A and B and C) = P(A)P(B)P(C).
So let’s apply the two new concepts to our online shopping example.
Back to the Shopping Online Example
花生的谜语怎么说Results for a population of 1000 customers that shopped online at a particular website during the past holiday ason. Recorded whether or not they were satisfied with the experience and whether or not they received the products on time.
g. Are being satisfied with the experience and receiving the product on time mutually exclusive
(disjoint )? Provide support for your answer.
< are there any customers in the interction, that were both satisfied and received it on
time? Since there are 800 customers in the interction, the two events are NOT MUTUALLY EXCLUSIVE (that is, NOT DISJOINT).
h. Are being satisfied with the experience and receiving the product on time statistically
independent ? Provide support for your answer.
Hint: go back and compare your answers to parts a, e, and f.
From (a) we have P(satisfied) = 0.82; from (b) we have P(satisfied | on time) = 0.901; and from part (f) we have P(satisfied | not on time) = 0.167. Since knowing the product was received on time did increa the probability of the customer being satisfied (from 0.82 to 0.901), the two events are NOT INDEPENDENT. Note: we could also check if P(Satisfied and On Time) = P(Satistifed)P(On Time). This is the multiplication rule for independent events. Since 0.80 does not equal (0.82)(0.88) = 0.7216, again the same conclusion is reached not independent.
Suppo that in a certain country, 10% of the elderly
people have diabetes. It is also known that 30% of the
elderly people are living below poverty level and 5% of the
笨组词elderly population falls into both of the categories. At the
right is a diagram for the events. Do the probabilities make
n to you?
a. What is the probability that a randomly lected elderly person is not diabetic? P(not diab) = 1 – P(diab) = 1 – 0.10 = 0.90 or 90%
b. What is the probability that a randomly lected elderly person is either diabetic or living below
poverty level?
P(diab or BP) = P(diab) + P(BP)- P(diab and BP) = 0.10 + 0.30 – 0.05 = 0.35 or 35%
c. Given a randomly lected elderly person is living below poverty level, what is the probability that
he or she has diabetes?
P(diab|BP)=P(diab and BP)=0.05= 0.167 P(BP) 0.30
d. Since knowing an elderly person lives below the poverty level (circle one) DOES DOES NOT
change the probability that they are diabetic, the two events of living below the poverty level and being diabetic (circle one) ARE ARE NOT independent.
