| << Chapter < Page | Chapter >> Page > |
When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not.
If A and B are two events defined on a sample space , then: P ( A AND B ) = P ( B ) P ( A | B ).
This rule may also be written as: P ( A | B ) =
(The probability of A given B equals the probability of A and B divided by the probability of B .)
If A and B are independent , then P ( A | B ) = P ( A ). Then P ( A AND B ) = P ( A | B ) P ( B ) becomes P ( A AND B ) = P ( A ) P ( B ).
If A and B are defined on a sample space, then: P ( A OR B ) = P ( A ) + P ( B ) - P ( A AND B ).
If A and B are mutually exclusive , then P ( A AND B ) = 0. Then P ( A OR B ) = P ( A ) + P ( B ) - P ( A AND B ) becomes P ( A OR B ) = P ( A ) + P ( B ).
Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska
Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game.
A = the event Carlos is successful on his first attempt.
P (
A ) = 0.65.
B = the event Carlos is successful on his second attempt.
P (
B ) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal
GIVEN that he made the first goal is 0.90.
a. What is the probability that he makes both goals?
a. The problem is asking you to find P ( A AND B ) = P ( B AND A ). Since P ( B | A ) = 0.90: P ( B AND A ) = P ( B | A ) P ( A ) = (0.90)(0.65) = 0.585
Carlos makes the first and second goals with probability 0.585.
b. What is the probability that Carlos makes either the first goal or the second goal?
b. The problem is asking you to find P ( A OR B ).
P ( A OR B ) = P ( A ) + P ( B ) - P ( A AND B ) = 0.65 + 0.65 - 0.585 = 0.715
Carlos makes either the first goal or the second goal with probability 0.715.
c. Are A and B independent?
c. No, they are not, because P ( B AND A ) = 0.585.
P ( B ) P ( A ) = (0.65)(0.65) = 0.423
0.423 ≠ 0.585 = P ( B AND A )
So,
P (
B AND
A ) is
not equal to
P (
B )
P (
A ).
d. Are A and B mutually exclusive?
d. No, they are not because P ( A and B ) = 0.585.
To be mutually exclusive, P ( A AND B ) must equal zero.
Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P ( C ) = 0.75. D = the event Helen makes the second shot. P ( D ) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws?
P ( D | C ) = 0.85
P (
C AND
D ) =
P (
D AND
C )
P (
D AND
C ) =
P (
D |
C )
P (
C ) = (0.85)(0.75) = 0.6375
Helen makes the first and second free throws with probability 0.6375.
A community swim team has
150 members.
Seventy-five of the members are advanced swimmers.
Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers.
Forty of the advanced swimmers practice four times a week.
Thirty of the intermediate swimmers practice four times a week.
Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.
a. What is the probability that the member is a novice swimmer?
a.
b. What is the probability that the member practices four times a week?
b.
c. What is the probability that the member is an advanced swimmer and practices four times a week?
c.
d. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not?
d.
P (advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.
e. Are being a novice swimmer and practicing four times a week independent events? Why or why not?
e. No, these are not independent events.
P (novice AND practices four times per week) = 0.0667
P (novice)
P (practices four times per week) = 0.0996
0.0667 ≠ 0.0996
Notification Switch
Would you like to follow the 'Introductory statistics' conversation and receive update notifications?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|