2.3 Notes/Examples

Sometimes there are situations in which two different outcomes cannot occur at the same time. For example, if you roll a single die one time and you wish to find the probability of getting an even number and a 3 on that one roll. These two outcomes cannot occur at the same time. When it is impossible for two outcomes to occur at the same time, we say the outcomes are mutually exclusive or disjoint. If outcomes 'A' and 'B' are mutually exclusive then it is impossible for outcome 'A' and 'B' to happen at the same time, or P(A and B)=0. However, if 'A' and 'B' are mutually exclusive then P(A or B)=P(A)+P(B). Using proper notation we have

$P(A \cup B)=P(A)+P(B)$

Remember, this is only true if the two outcomes are mutually exclusive.

The U can be read as the union of outcomes 'A' and 'B'. The probability for the union of two outcomes 'A' and 'B' can be thought of as the chance that either 'A' occurs, 'B' occurs, or both 'A' and 'B' occur.

Example 1

A single die is rolled one time. What is the probability of getting either an odd number or a 6?

Solution

The outcomes of getting an odd number and getting a 6 are mutually exclusive since they cannot occur at the same time.

$P(Odd\cup6)=P(Odd)+P(6)=\frac{3}{6}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}\approx0.67$

Of course, not every situation involves mutually exclusive events.

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$

[Figure3]

The diagrams in Figure 2.1 on the following page are Venn diagrams. They are useful in showing us how different outcomes are related. Outcomes 'A' and 'B' are not mutually exclusive if they overlap and we can say that there is an intersection where outcomes 'A' and 'B' overlap.

For example, if we roll a single 6-sided die, the outcomes of getting an odd number and getting a number bigger than 3 intersect. They both include the number 5. The symbol for an intersection is ∩

A logical extension of the formula for the union of two outcomes that are not mutually exclusive is...

[Figure4]

Example 2

A single 6-sided die is rolled. Suppose the outcomes we are interested in are getting an odd number and getting a prime number (2,3, or 5). Draw a Venn diagram for this situation.

Solution

Notice that since the numbers 3 & 5 belong to both the odd numbers and the prime numbers, they are placed into the intersection of the 'Odds' circle and the 'Primes' circle. Notice also that the numbers 4 & 6 do not belong to either set and are placed outside both circles.

Example 3

A single 6-sided die is rolled. What is the probability of getting either an odd number or a prime number? Note that this is the same as asking for

$P(Odd \cup Prime)$

Solution

$\frac{4}{6}=\frac{2}{3}\approx0.67$

Using the figure from Example 1 we see that there are four values out of six that are either odd or prime. Therefore, P(Odd or Prime)=. If we use the formula,

$\frac{3}{6}+\frac{3}{6}-\frac{2}{6}=\frac{4}{6}=\frac{2}{3}\approx0.67$

$P(Odd \cup Prime)=P(Odd)+P(Prime)-P(Odd \cap Prime)$

. This gives .

Example 4

Suppose there is a 60% chance it will rain today and that there is a 70% chance that it will be over 90°F. Suppose also that there is a 45% chance that it will both rain and be above 90 degrees. What is the chance that it will neither rain nor be above 90 degrees? Solve using both a Venn diagram and using a formula.

Solution

Start by drawing a Venn diagram and noting that we have two circles, one for rain and one temperature. These two outcomes are not mutually exclusive because they can occur at the same time therefore the circles should overlap. We can quickly fill in the intersection of the two outcomes as 45%.

If we remember that there is a 60% chance of rain and we already have 45% filled in for the rain circle, the remainder of the rain circle must be 15% so it adds up to 60%. Likewise, the remainder of the >90° circle must be 25%.

We can now see that we have a total of 15%+45%+25%=85%. This means that the 'Neither' category must be 15% to give us a total of 100%.

Using the formula,

$P(Rain \cup 90)=P(Rain)+P(90)-P(Rain \cap 90)$

$P(Rain \cup 90)=60\%+70\%-45\%=85\%$

. Filling in we get . 100%-85%=15%.

Example 5

Which pairs of outcomes are mutually exclusive?

a) You go to the pet store to buy a pet. Outcome A = You buy a pet that flies, Outcome B = You buy a pet that has no legs.

b) You order a pizza. Outcome A = Your pizza has pepperoni on it, Outcome B = Your pizza has mushrooms.

c) You select a football player to take a picture for the yearbook. Outcome A = The player is a 4-year varsity starter, Outcome B = The player is 14 years old.

d) Radio stations have 4-letter station names such as KDWB. You decide to pick a radio station to listen to. Outcome A = The station's 4-letter name starts with a W, Outcome B = The station's 4-letter name contains three E's.

Solution

a) Is mutually exclusive. You cannot buy a pet that both flies and also has no legs.

b) Is not mutually exclusive. You can order a pizza with both mushrooms and pepperoni.

c) Is mutually exclusive. If a 4-year varsity starter is 14 years old, then they would have been a varsity starter when they were 10 years old. That simply does not happen.

d) Is not mutually exclusive. These both could happen if the radio station's call sign was WEEE.

Example 6

The Rockin' Rollers performance company has 30 musicians who play either bass guitar, lead guitar, or rhythm guitar. Some of these musicians play more than one instrument. Suppose 4 musicians can play lead, rhythm, or bass guitar. Fourteen can play lead or rhythm but not bass, two can play bass or rhythm but not lead, 3 can play bass only, and 4 can play rhythm only. There are no musicians who play lead and bass only. Draw a Venn diagram to determine how many musicians play lead only.

[Figure10]

Solution

From the information given, we can fill in a Venn diagram as shown below.

We've used up 27 of the 30 musicians so there must be 3 musicians who play lead guitar only. [Figure11]

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