# Elementary Statistics Triola, Elementary Statistics 11/e Unit 4 The

```Elementary Statistics
Triola, Elementary Statistics 11/e
Unit 4 The Multiplicative Rule
When we talk about event A occurring and event B occurring, things start to get a bit more complicated.
We can either be talking about the two events occurring at exactly the same time, or first one event
occurs and then the other occurs.
Consider tossing a coin. The sample space is {H,T}. It is not possible to toss the coin once and get both
heads and tails. These are mutually exclusive or disjoint events and cannot happen at the same time.
Therefore,
𝑃(𝐻 ∩ 𝑇) = 0
Now, let’s change things up a bit, and say that we toss the coin twice. Now the sample space is {HH, HT,
TH, TT). If order doesn’t matter, just that we get a heads and a tails, we have,
𝑃(𝐻 ∩ 𝑇) = 1/2
So, you see, context makes a big difference. In the weeks ahead you are going to be asked questions
that will be long and wordy. It will be necessary for you to read the entire question carefully before
Independent Events
Event A and event B are said to be independent if the probability of event B occurring depends in no
way on whether or not event A has occurred. Once again consider the case of tossing a coin twice. If
event A is defined to be getting a heads on the first toss, and event B is getting a heads on the second
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toss, then 𝑃(𝐵) = 2
no matter what happens on the first toss, because the two events are
independent.
Condtional Events
An event B is said to be conditional on event A if the probability of B occurring depends in some way on
whether or not A has occurred. Another term used when one event is conditional on another is to say
that the events are dependent. A and B are dependent if the probability of one occurring depends on
whether or not the other occurred as well. Consider the following example.
Two canoes are coming down a river, one following behind the other, when they encounter a rock dam.
There is only one narrow gap in the dam where they can pass without getting caught up on the rocks,
but the gap is impossible to see from a canoe. Therefore, it is purely random whether or not the first
canoe makes it through. However, the experience of the first canoe is going to affect the decisions
made by the people in the second canoe, regardless of whether or not the first one makes it through.
Therefore, the probability of the second canoe making it through the rock dam depends on what
happens to the first canoe. Hence, its probability of making it through the dam is conditional on the first
canoe.
We denote the probability of ‘B conditional on A’ as 𝑃(𝐵|𝐴).
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Unit 4
The Multiplicative Rule
Multplicative Rule
In general, the probability of A and B occurring is given by,
𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) ∙ 𝑃(𝐵|𝐴)
Let’s see this rule at work. Take another look at the probability table in Unit 2 involving the testing of
marijuana users. Let A be then event that the selected person tested positive and let B be the event
that the person selected uses marijuana. We saw the following,
𝑃(𝐴) =
143
300
𝑃(𝐴 ∩ 𝐵) =
119
300
Now, let’s think about P(B|A). Given A means that we know the person selected tested positive. Let’s
take all the people in the room that tested positive and put them in a room by themselves. That way,
when we select a person at random from that room, we’re sure to know that they had tested positive.
There will be 143 people in that room. Now, what is the probability of selecting someone from that
room that is a user. In this room, there will be exactly 119 people who are users. Look at the table and
think about this. Hence, the probability of selecting a user from that room is 119 out of 143, but this is
exactly the probability of event B occurring, i.e. we selected a marijuana user given that the person had
tested positive. Therefore,
𝑃(𝐵|𝐴) =
119
143
𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵|𝐴) =
and
143 119 119
∙
=
300 143 300
Question 1
Using the table in Unit 2, let A be the event that the selected person did not test positive and let B be
the event that the selected person was not a user. Find P(A), P(B|A), P(A)P(B|A) and P(A∩B).
As a corollary, if we know P(A∩B) and P(A) we can find P(B|A),
𝑃(𝐵|𝐴) =
This is the end of Unit 4.
𝑃(𝐴 ∩ 𝐵)
(𝐴)
Turn now to your MyMathLab homework to get more
practice with these concepts.
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