Sec. 7.3 - Probability of Independent Events
In this section, we will look at the chance (probability) of two or
more events occurring together. The events will be independent
events - that is, events which do not affect each other.
Some definitions:
Probability
o The chance that some event(s) will occur. For example,
the probability of rolling a number less than 5 on a
regular six-sided die is
4
2
, which simplifies to .
6
3
 We can write this as P(number less than 5) =
2
.
3
Independent events – events which do not affect each other.
Examples – which events are independent?
1)
Event A - Flip a coin.
Event B - Roll a die.
Is the outcome on the die affected by the outcome on
the coin?
2)
Three red marbles and four blue marbles are placed in
a bag. A marble is randomly selected, the color is
recorded, and the marble is placed back in the bag
before a second selection is made.
Is the outcome on the second selection affected
by the outcome on the first selection?
3)
Three red marbles and four blue marbles are placed in a
bag. A marble is randomly selected, the color is recorded,
but the marble is NOT placed back in the bag before a
second selection is made.
Is the outcome on the second selection affected by the
outcome on the first selection?
The above examples are meant to help you understand the
meaning of independent events by comparing them with
dependent events. However, in this section we will be working
only with probabilities for independent events.
The Probability of Two Independent Events Occurring Together
Example:
A coin is flipped and a regular six-sided die is rolled. What is the
probability of obtaining a head on the coin and a 3 on the die –
that is, what is P(H and 3)?
Solution
There are several ways to look at this situation.
Method 1: Tree Diagram
List the possible outcomes in a tree diagram:
Outcomes on Coin
Outcomes on Die
Possible
Outcomes
Only one of the 12 possible outcomes is the desired outcome.
Therefore, P(H and 3) =
1
.
12
Method 2: Use a table. Record the possible outcomes.
Roll of Die
Toss
Coin
1
2
3
4
5
6
H
T
Only one of the 12 possible outcomes is the desired outcome.
Therefore, P(H and 3) =
1
.
12
Method 3: Find the probability of each individual event and then
multiply the probabilities.
P(H) =
1
2
P(3) =
1
6
Therefore, P(H and 3) = P(H) × P(3)
=
1 1
2 6
=
1
12
Example:
Find the probabilities of obtaining a sum of 7 and a sum of 2 on a
roll of two dice.
We can list the possible outcomes in a table:
nd
2
1
1
1st die
2
3
4
5
6
Therefore:
P(sum of 7) =
P(sum of 2) =
2
Die
3
4
5
6
Example
30 white marbles and 15 blue marbles are placed in a bag. A
marble is randomly selected, placed back in the bag, and then a
second marble is selected.
Event W – select a white marble.
Event B – select a blue marble.
Find each of the following probabilities:
A)
P(W and B)
B)
P(W and W)
C)
P(B and not B)
For Further examples, see pp.408-410 in your textbook.