In this chapter, we’ll learn about the basic rules of probability, what it means for events to be independent, and about discrete and continuous random variables, simulation, and rules for combining random variables.
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Example:
If we roll a six-sided die, we know that we will get a 1, 2, 3, 4, 5, or 6, but we don’t know which one of these we will get on the next trial. What are two ways we could estimate the probability of rolling a 6?
Solution:
We could estimate the probability experimentally by rolling a die 300 times. The results of one such set of rolls are shown below.
Out of 300 trials, 56 were successes.
Thus, ==probability = 56/300 = 0.187==
We could also estimate the probability theoretically by assuming each possible result is equally likely.
If that is the case, then the probability of each number appearing = ==1/6 = 0.167==
Example:
Example:
Consider the experiment of flipping two coins. ==Sample space is HH, HT, TH, TT.== The list is complete because there is no other possibility for tossing two coins. The list is disjoint because a pair of coin tosses can result in only one of those outcomes.
If we let E = the event of interest, then the probability of E, written P(E) is given by:
The sum of the probabilities of all possible outcomes in a sample space is 1.
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Example 1:
The table below represents country of origin and blood type for 500 people attending a large conference.
If we were to randomly select one person from this population of people attending the conference, how would you verify that the outcomes in the table define a sample space?
Solution:
You could check that each person falls into exactly one category in the table.
For this, check that the row and column totals are actually the sum of the numbers in the cells.
If they are not, it would mean that either some people fall into more than one category or there would be people not included in the table.
==In this table, the totals are actually the sum of the numbers in the cell. This means the list is disjoint and complete.==
Example 2:
Using the data above, if you randomly select a person from the population of people attending the conference, what is the probability the person has blood type A?
Solution:
There is a 34% probability that a randomly selected person attending this conference has blood type A.
==P (A or B) = P (A) + P (B) – P(A and B)==
Example:
If you randomly select one person from the population from the table used previously, use the addition rule to find P(Type A or Country #2).
Solution:
P(Type A or Country #2) = P (Type A) + P (Country #2) – P (Type A and Country #2)
Example:
If you randomly select a person attending this conference, what is the probability that the person has blood type A, O, or AB?
Solution:
You could simply subtract the marginal probability for the one remaining category, Type B, and subtract that from 1.
==P (A, O, or AB) = 1 − P (Type B)==
Example (Conditional Probability from a table):
If you randomly select a person with blood type A, what is the probability that this person is from Country 3?
Solution:
You know the person has blood type A. We need to look only at the relevant column.
==P (Country 3 | Type A) = 22/170 = 0.129==
Example (Conditional Probability from a tree):
Suppose a computer company has manufacturing plants in three states. Fifty percent of its computers are manufactured in California, and 85% of these are desktops; 30% of computers are manufactured in Washington, and 40% of these are laptops; and 20% of computers are manufactured in Oregon, and 40% of these are desktops. All computers are first shipped to a distribution site before being sent to stores.
a)If you picked a computer at random from the distribution center, what is the probability that it is a laptop?
b) What are the chances that if you randomly selected a laptop from the distribution center, it is a laptop that was manufactured in California?
Solution:
a) Sum of all probabilities = 1
Now, ==P (laptop) = 0.075 + 0.12 + 0.12 = 0.315.==
b) ==P (California | laptop) = 0.075/0.315) = 0.238.==
Example:
Consider drawing one card from a standard deck of 52 playing cards.
i)Are A and B independent?
ii)Are A and C independent?
Solution:
i) P (A|B) =
P (the card drawn is an ace | the card is a 10, J, Q, K, or A)
= 4/20
= 1/5.
Since P (A) = 1/13, knowledge of B has changed what we know about A.
In this case, P (A) ≠ P (A|B), so events A and B are not independent.
ii) P (A|C) =
P (the card drawn is an ace | the card drawn is a diamond)
= 1/13
In this case, P (A) = P (A | C) so that the events “the card drawn is an ace” and “the card drawn is a diamond” are independent.
==P (A and B) = P (A) . P (B|A)==
If events A and B are independent,
P (B|A) = P(B)
==so P (A and B) = P (A). P (B)==
Example:
A and B are two mutually exclusive events for which P (A) = 0.3 and P (B) = 0.25. Find P (A or B). Solution: P (A or B) = 0.3 + 0.25 = 0.55.
A segmented bar graph takes bars of equal length and equal width for each of the groups and divides them up into segments that represent the percentage for each of the categories.
We use conditional relative frequencies
Looking at the previously used table:
To find relative frequencies, we take each country as its own group and calculate based on that condition.
For example, there are 150 people out of 250 in Country 1 with blood type O, and this would be a conditional relative frequency of 0.60 or 60%.
Question: Based on this segmented bar graph, do you believe that blood type and country are independent?
Answer: Since the segments are so differently sized in each bar, it does not appear that blood type and country are independent.
Another type of plot that can be used with these data is a mosaic plot.
A mosaic plot helps preserve the relative sizes of the groups by keeping the heights the same but making the widths of the bars proportional to the group size.
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Example:
If we roll a fair die, the random variable X could be the face-up value of the die. The possible values of X are {1, 2, 3, 4, 5, 6}. P ( X = 2) = 1/6.
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Example:
Let X be the number of boys in a three-child family. Assuming that the probability of a boy on any one birth is 0.5, the probability distribution for X is
==The probabilities Pi of a DRV satisfy two conditions:==
The mean of a discrete random variable, also called the expected value , is given by
The variance of a discrete random variable is given by
The standard deviation of a discrete random variable is given by
The following is a TI-83/84 probability histogram for discrete variables of the probability distribution we used in a couple of the earlier examples.
For a probability Distribution for a Continuous Random Variable, Here are a few things to remember:
Example:
In a standard normal distribution, what is the probability that z < 1.5?
Solution:
The standard normal table gives areas to the left of a specified z -score. From the table, we determine that the area to the left of z = 1.5 is 0.9332.
Example:
A coin is known to be biased in such a way that the probability of getting a head is 0.4. If the coin is flipped 50 times, how many heads would you expect to get?
Solution:
Let 0, 1, 2, 3 be a head and 4, 5, 6, 7, 8, 9 be a tail. If we look at 50 digits beginning with the first row, we see that there are 18 heads, so the proportion of heads is ==18/50 = 0.36==. This is close to the expected value of ==0.4.==
Example:
One contractor can finish a particular job, on average, in 40 hours ( μx = 40). Another contractor can finish a similar job in 35 hours ( μy = 35). If they work on two separate jobs, how many hours, on average, will they bill for completing both jobs?
Solution:
It should be clear that the average of X + Y is just the average for X plus the average for Y. That is, ==μX ± Y = μX ± μY .==
To combine the variances, we use:
To combine standard deviations,
Example:
A private school offers an admission test on the first Saturday of November and the first Saturday of December each year. In 2008, the mean score for hopeful students taking the test in November was 156 with a standard deviation of 12. For those taking the test in December, the mean score was 165 with a standard deviation of 11. What are the mean and the standard deviation of the total score X + Y of all students who took the test in 2008?
Solution:
It is reasonable to assume that X and Y are independent. Therefore:
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