Complements and Conditional Probability in Statistics
These are my notes on complements and conditional probability.
Complements
When finding the probability of some event occurring at least once, we should understand that at least one has the same meaning as one or more. The complement of getting at least one particular event is that you get no occurrences of that event.
Finding the probability of getting at least one of some event:
- Let A = getting at least one of some event.
- Then \(/bar{A}\) = getting none of the event being considered.
- Find \(P(/bar{A})\) = probability that event A does not occur.
- Subtract the result from 1.
Conditional Probability
A conditional probability of an event is a probability obtained with the additional information that some other event has already occurred.
\[P(B | A)\] denotes the conditional probability of event B occurring, given that event A has already occurred.
Intuitive Approach For Finding P(B|A)
The conditional probability of B occurring given that A has occurred can be found by assuming that event A has occurred and then calculating the probability that event B will occur.
Formal Approach For Finding P(B|A)
The probability P(B|A) can be found by dividing the probability of events A and B both occurring by the probability of event A.
\[P(B|A)=\frac{P(A \text{and} B)}{P(A)}\]
The preceding formula is a formal expression of conditional probability, but blind use of formulas is not recommended. The intuitive approach is recommended.
Bayes’ Theorem
The importance and usefulness of bayes’ Theorem is that it can be used with sequential events, whereby new additional information is obtained for a subsequent event, and that new information is used to revise the probability of the initial event. In this context, the terms prior probability and posterior probability are commonly used.
A prior probability is an initial probability value originally obtained before any additional information is obtained.
A posterior probability is a probability value that has been revised by using additional information that is later obtained
Multiplication Counting Rule
The multiplication counting rule is used to find the total number of possibilities from some sequence of events. For a sequence of events in which the first event can occur n ways, the second event can occur n2 ways and so on, the total number of outcomes is n1*n2*n3….
Factorial Rule
The factorial rule is used to find the total number of ways that n different items can be rearranged. The factorial rule uses the following notation. The factorial symbol(!) denotes the product of decreasing positive whole numbers. The factorial rule is stated as the number of different arrangements of n different items when all n of them are selected is n! The factorial rule is based on the principle that the first item may be selected n different ways, the second item may be selected n-1 ways, and so on. This rule is really the multiplication counting rule modified for the elimination of one item on each selection.
Permutations and Combinations
When using different counting methods, it is essential to know whether different arrangements of the same items are counted only once or are counted separately. The terms permutations and combinations are standard in this context.
Permutations of items are arrangements in which different sequences of the same items are counted separately.
Combinations of items are arrangements in which different sequences of the same items are counted as being the same.
Permutations Rule
The permutation rule is used when there are n different items available for selection, we must select r of them without replacement, and the sequence of the items matters. The result is the total number of arrangements that are possible. Remember, rearrangements of the same items are counted as different permutations.
\[nP_r=\frac{n!}{(n-r)!}\]
When n items are all selected without replacement, but some items are identical, the number of possible permutations is found by using the following rule:
\[\frac{n!}{n_1!n_2!...n_k!}\]
Combinations Rule
The combinations rule is used when there are n different items available for selection, only r of them are selected without replacement, and order does not matter. The result is the total number of combinations that are possible. Remember, rearrangements of the same items are considered to be the same combination.
\[n_C_r=\frac{n!}{(n-r)!r!}\]
Find the probability that when a couple has three children, at least one of them is a girl. Assume that boys and girls are equally likely.
For each event there are two possibilities. There are 3 events.
½*½*½ = ⅛
1-⅛=⅞
In a certain country, the true probability of a baby being a girl is .509. Among the next six randomly selected births in the country, what is the probability that at least one of them is a girl?
The probability of at least one can be computed using the rule of complements. Let A represent the event that at least one of the next six births is a girl. Use the rule of complements below to find the probability of event A, P(A), where \(\bar{A}\) is the complement of A.
\[P(A)=1-P(\bar{A})\]
The complement of A, \(\bar{A}), is the event that the next six births are all boys.
Since each birth has no effect on any of the other births, the births are all independent events. The probability that the next six births are all boys can be found using the multiplication rule for independent events. The probability of the event can be written as shown below:
It is given that the probability of a birth being a boy is .509.
Use the multiplication rule for independent events to find the probability that the next six births are all boys. The multiplication rule for independent events states that the probability of two independent events occurring is the product of their individual probabilities. This can be extended to 6 independent events.
.509*.509*.509*.509*.509*.509 = .017
Then use the rule of complements to find the probability that the couple has at least one girl.
1-.017=.983
Therefore, the probability that the next six randomly selected births will contain at least one girl is .983
Subjects for the next presidential election poll are contacted using telephone numbers in which the last four digits are randomly selected (with replacement). Find the probability that for one such phone number, the last four digits include at least one 0.
10^4-9^4=3439
10^4=10000
3439/1000=.344
Based on a poll, 72% of internet users are more careful about personal information when using a public wi-fi hotspot. What is the probability that among three randomly selected internet users, at least one is more careful about personal information when using a public wi-fi hotspot? How is the result affected by the additional information that the survey subjects volunteered to respond to?
The probability of at least one can be computed using the rule of complements. The rule of complements states that the following expression is true for events A and \(\bar{A}\), where \(\bar{A}\) indicates that event A did not take place.
\[P(A)=1-P(\bar{A})\]
Identify the event that is the complement of A
\[\bar{A} = \text{none of the internet users are more careful}\]
To find the probability of the complement, first find the probability that an internet user is not more careful with personal information while using a public wi-fi hotspot.
1-P(is more careful)
1-.072 = .28
Find the probability of the complement using the multiplication rule for independent events, rounding to three decimal places. The multiplication rule for independent events states that the probability of two independent events occurring is the product of their individual probabilities. This can be extended to three independent events.
.28 * .28 * .28 = .022
Now use the rule of complements
1-.022 = .978
It is very possible that this result is not representative of people that use wi-fi
In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it on gum.
Purchased Gum Kept the Money
Given four quarters 37 13
Given $1 bill 11 39
- Find the probability of randomly selecting a student who spent the money, given that the student was given four quarters.
The conditional probability of B occurring given that A has occurred, P(B|A), can be found intuitively by assuming that event A has occurred and then calculating the probability that event B will occur.
More formally, the probability P(B|A) can be found by dividing the probability of events A and B both occurring by the probability of event A.
In this case, given four quarters corresponds to event A and spent the money corresponds to event B.
First determine the number of students given four quarters that spent the money
37 students
Now calculate the probability
37/50=.74
- Find the probability of randomly selecting a student who kept the money given that the student was given four quarters.
Recall that 50 students were given four quarters
Identify the number of students given four quarters that kept the money
13 students
Now calculate the probability
13/50=.26
Now that since the students either kept the money or spent the money, these probabilities are complements.
.26=1-.74
- What do the preceding results suggest?
Compare the probabilities found in first parts
Spent the money=.74
Kept the money=.26
Since .74..26 P(spent the money | four quarters) has the greater probability
The accompanying table shows the results from a test for a certain disease. Find the probability of selecting a subject with a negative test result, given that the subject has the disease. What would be an unfavorable consequence for this error?
357 26
18 1150
A conditional probability of an event is a probability obtained with the additional information that some other event has already occurred. P(B|A) denotes the conditional probability of event B occurring, given that event A has already occurred.
A is the event which is known to have occurred. The given event is “the individual has the disease”.
B is the event for which the probability is sought. The event is “the individual tests negative for the disease”.
The conditional probability of B given A can be found by assuming that event A has occurred and, working under that assumption, calculating the probability that event B will occur.
First, determine the number of individuals who have the disease. Add all the values in the indicated column.
357+18=375
From the table, there are 18 individuals who have the disease and test negative. Divide to find the probability
18/375=.048
Therefore, the probability that a randomly selected individual who has the disease tests negative is.048
To determine an unfavorable consequence of this error, consider a subject that has the disease but with a negative test result.
Note that negative test results would lead the subject to believe that they have the disease.
The table below displays results from experiments with polygraph instruments. Find the positive predictive value for the test. That is, find the probability that the subject lied, given that the test yields a positive result.
Did not lie lied
Pos 9 46
Neg 30 13
Use the intuitive approach to conditional probability. The conditional probability of B occurring given that A has occurred can be found by assuming that event A has occurred and then calculating the probability that event B will occur. Find the probability of selecting a subject who lied, given that the selected subject had a positive test result. If it is assumed that the subject had a positive test result, then only the 9+46=55 subjects in the top row of the table are to be used. Among those 55 subjects, 46 subjects who had a negative test result actually lied.
Divide the number of subjects who had a positive test result and actually lied by the total number of subjects who had a positive test result to find the probability, rounding to three decimal places.
46/55=.836
Assume that there is a 12% rate of disk drive failure in a year.
- If all your computer data is stored on a hard disk with a copy stored on a second hard disk drive, what is the probability that during a year, you can void catastrophe with at least one working drive.
- If copies of all your computer data are stored on three independent disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive.
- Use the rule of complements shown below to find the probability that you can avoid catastrophe. Let A=at least one hard drive works correctly
\[P(A)=1-P(bar{A})\]
Identify the event that is the complement of A
\[\bar{A}=both hard drives fail\]
Since the two hard drives operate separately, their failures are independent events. Use the multiplication rule for independent events to find the probability of the complement of event A. The multiplication rule for independent events states that the probability of two independent events occurring is the product of their individual probabilities. The probability of any one of the hard drives failing to work correctly is 0.12
.\[ P(\bar{A})= .12*.12=.0144 \]
Now find P(A) by evaluating \(1-P(\bar{A})\)
1-.0144 = .9856
- Again, let A = at least one hard drive works correctly.
\[P(\bar{A}) = .12 * .12 * .12 = .001728 \]
Now find P(A) by evaluating \(1-P(\bar{A})\)
1-.001728 = .998272