Section P5: Conditional Probability
In some circumstances, having information about an outcome can change the probability of a particular
event occurring.
Example:
If a fair die is rolled the probability of rolling a 5 is 1/6. However, if I am told that the number rolled is
an odd number, the possible outcomes are 1,3,5. The sample space is now limited to only 3 outcomes,
so the probability of rolling a 5 is 1/3. By limiting the possible number of outcomes, the probability of
the event may be changed.
The probability that an event E will occur given that the event F has occurred, is called a conditional
probability and is denoted by P(E/F). This is read “the probability of E given F.”
Usually, the condition will limit the sample space to a subset of what it would otherwise have been.
Example:
Each year the local library adds a number of new publications. The chart below categorizes them as
hardcover or paperback and fiction or nonfiction.
Fiction (F)
Nonfiction (N)
Totals
12
15
27
18
10
28
30
25
55
Hardcover (H)
Paperback (P)
Totals
If one of the 55 books is randomly chosen for reading, find the probability it will be a hardcover.
Since 30 of the 55 books are hardcover, P(H) = 30/55 = 6/11.
Find the probability that the book is fiction, given it is hardcover: Only 30 of the books are hardcover,
so this reduces the sample size to 30. In addition, only 12 of the fiction books (the event) are hardcover
so this reduces the event size: P( F/H) = 12 /30 = 2/5.
Notice that the number in fiction and hardcover is actually the number in the numerator of the
probability. This leads to the general rule for conditional probability.
Conditional Probability
If E and F are two events, then the probability of E given the event F can be found by:
P(E/F) =
Example:
Two fair dice are rolled. Find the probability that the sum is 8, given that both dice show the same
number.
1
E is the event of getting a sum of 8, F is the event of having both numbers the same.
So the P(E/F) =
The sample space is limited by the given part (F) to only (1,1)(2,2)(3,3)(4,4) (5,5)(6,6).
So n(F) = 6
The event E of getting a sum of 8 and the event F of numbers being the same is the overlap of the two
events. Only (4,4) has a sum of 8 and has both numbers the same. So n(E∩F) = 1
The Probability of E given F is: P(E/F) = 1/6.
Sometimes we do not know the exact counts of n(E and F) and n(F). If we know P(E and F) and P(F),
we can still compute the conditional probability, because:
P(E/F) =
.
So the alternate formula for conditional probability is:
P(E/F)=
.
Example: Suppose a poll of local voters indicates that 42% favor a property tax increase to support
schools, 60% favor adding horse racing as a means of creating state revenue, and 15% favor both. If a
voter is in favor of the property tax increase, what is the probability that the voter is also in favor of the
horse racing?
We have probabilities here, so we will use the alternate formula.
The information we are given is that voter is in favor of the property tax increase, so that is event F.
P(F) = .42
The event E is being in favor of horse racing, so the overlap of favoring both the property tax and horse
racing is P(E and F)= .15
So the P(E/F) =
≈ 0.3571 = 35.71%
The Multiplication Rule for Dependent Events
Suppose we want to find the probability of both event E and F occurring. We can use the rule for
conditional probability to find a formula for this.
P(E/F)=
. We can solve this equation for the P(E and F) and get what is called the
multiplication rule.
P(E/F)=
; so P(E∩F) = P(F)∙P(E/F).
This rule is used when what happens in event E is dependent on what happens in event F. In other
words, the probability of E and F occurring is affected by the occurrence of the event F.
2
Example:
Two cards are drawn from a deck of cards without replacement. What is the probability that both cards
are aces?
Since the cards are not replaced we must use the multiplication rule for dependent events.
The event of the first card being an ace is the event F. P(F) =
an ace, given that the first card is an ace is E/F. P(E/F) =
. The event of the second card being
.
So P(first card an ace and second card an ace) = P(F∩E) = P(first card is an ace)∙P(second card is an ace,
given first is an ace) = P(F) ∙P(E/F)=
.
Example:
Meteorologists have difficulties accurately predicting when it will rain, especially in a short time period.
They appear to be more accurate predicting when it will not rain. The following table gives the
outcomes of 1000 one-hour timeframes based on the prediction of rain or no rain.
Forecast rain
Forecast no rain
Total
rain
no rain
Total
62
158
220
18
762
780
80
920
1000
1) What is the probability that rain was forecast and it rained?
2) What is the probability that no rain was forecast and it did not rain?
1) Rain being forecast is the given event: P(rain forecast) = P(F) = 220/1000
Forecast correct and rain forecast is what we are looking for. (E∩F)
Rain given rain predicted is E/F. P(E/F) = 62/220
So P(E∩F)= P(F)∙P(E/F) =
=
= 0.062 or 6.2% of the time.
2) No rain being forecast is the given event: P(F) = 780/1000
Forecast correct and no rain forecast is what we are looking for, (E∩F).
No rain given no rain is predicted is E/F. P(E/F) = 762/780.
So P(E∩F)= P(F)∙P(E/F) =
=
= 0.762 or 76.2% of the time.
Notice that this information can be gleaned directly from the table: 1) 62 is the value in the box “forecast
rain and rain.” Forecast rain and rain is what we are looking for, so if we use that value as n(E∩F) and
the total sample space as n(S) we get: P(forecast rain and rain) = P (E∩F) = 62/1000.
This will work if you have a table, but you may not be given a table in a particular problem.
HW P5a, # 32: See handout
3
Independent Events
Sometimes the occurrence of an event F, does not affect the likelihood of an event E occurring. If that is
the case then, E and F are said to be independent events.
Two events, E and F are called independent events if knowledge about the occurrence of one of them
has no effect on the probability of the other one, thus: P(E/F) = P(E).
In this case we can use the probability of E and F occurring can be reduced to this formula:
P(E∩F) = P(E)∙P(F), if E and F are independent.
Example:
Two cards are drawn from a deck of 52 cards, replacing the first card after it is drawn. What is the
probability that both cards are aces?
Since the first card is replaced after it is drawn, the two events of an ace on the first draw, E, and an ace
on the second draw, F, are independent.
So P(E∩F) = P(E)∙P(F) =
=
∙
.
Example:
A grocery warehouse encourages customers to buy in bulk. On any given day, 80% of the customer
sales amount to more than $125. In other words, any given sale has a probability of .80 of being for
more than $125. (Assume that the first several sales of the day can be treated as independent.)
1) What is the probability that the first two sales on Monday are both for more than $125?
P(1st sale > 125 and 2nd sale > 125) = P (1st sale >125) ∙ P(2nd sale > 125) = 0.8 ∙0.8 = 0.64
2) What is the probability that exactly one of the first three sales is for more than $125?
First we must realize that there are C(3,1) or 3 ways that one of the first three sales could be more than
$125. Either the 1st sale or the 2nd sale or the 3rd sale could be the one over $125.
So to find the probability we have two parts, first decide how many ways we can have one sale greater
than $125 out of 3 sales, and then find the probability of one greater then $125 and two not greater than
$125.
We have for the first part: 3 ways
The second part: P(one sale > 125 and one sale < 125 and one sale < 125) = 0.8∙0.2∙0.2 = 0.032
Now by the fundamental theory of counting we multiply the two results together: 3 ∙ 0.032 = 0.096.
Example:
If a coin is flipped and a die is tossed , what is the probability of getting a tail and a 4?
These events are independent so P(tail and 4) = P(tail)∙P(4) =
4
Example:
Five dice are rolled.
1) What is the probability of getting exactly three fours?
2) What is the probability of getting at least 3 fours?
1) This is a problem involving two steps. First we must decide which three of the 5 dice have a four
showing: C(5,3) = 10 ways this can be done.
And then we must find the probability of three fours and two numbers not fours. So P(four and four and
four and not four and not four) = P(4)∙P(4)∙P(4)∙P(not 4)∙P(not 4) =
.
Now use the fundamental principle of counting to find the P(exactly three fours) =
10∙
Try part 2)
(HINT) To find at least three fours, we need to find three fours or four fours or five fours.
Do this in the same way as we found part 1 and then add all three results (because they are separated by
the word or).
P(at least three fours) = P(3 fours) + P(4 fours) + P(5 fours) =
+
=
=
This is an example of Binomial Probability.
An experiment that consists of repeated independent trials (the same thing repeated over and over) with
only two outcomes that can be defined as success or failure, is called a binomial experiment. If the
probability of success is called p and the probability of failure is 1- p, the probability of r successes in n
trials is given by:
C(n,r)pr(1-p)n-r .
The same exact experiment must be repeated over and over and the outcomes must be able to be defined
as success or failure to use this formula.
Example:
A baseball player has a well-established career batting average of .300. In a brief series with a rival
team, Scott will bat 10 times. Find the probability he will get more than two hits in the series?
It is easier to use the complement, E’: getting two or less hits.
This is a binomial experiment that is repeated 10 times. P(success) = .3 and P(failure) = .7
To find E’ we need to find P(0 hits) + P(1 hit) + P(2 hits)
0 hits: C(10,0) (.3)0 (.7)10 ≈ 0.0282
1 hit: C(10,1) (.3)1(.7)9 ≈ 0.1211
2 hits: C(10,2) (.3)2 (.7)8 ≈ 0.2335
So P(E’) = 0.0282 + 0.1211 + 0.2335 = 0.3828
P(E) = 1 – 0.3828 = 0.6172, so he has a 61.72% chance of getting more than two hits.
5
Try: In a certain state, it has been shown that only 50% of the high school graduates who are capable of
college work actually enroll in colleges. Find the probability that, among nine capable high school
graduates in the state, a) exactly four will enroll; b) from 4 through 6 will enroll.
These are binomial probabilities. (Ans: 0.246; 0.656)
HW P5b, # 33: See handout
6