Probability Unit
Probability Day 6 Notes: Conditional Probability, Day 2
I. Review
1.
Sixty-three random people were asked their gender and if they wear their seatbelts regularly. The table
shows the results of this survey. If a random person was
Yes
No
Total
selected, find the probabilities below.
Male
16
12
28
a. P (male)
28 4
Female
20
15
35
  0.4444
Total
36
27
63
63
9
b. P(wears seatbelt regularly)
c. P(male
d. P(male
36 4
  0.5714
63 7
 does not wear seatbelt regularly) 11  0.1905
63
 does not wear seatbelt regularly) P( M )  P( SB)  P( M  SB)
e. P(male | does not wear seatbelt regularly)
28 27 12 43



 0.6825
63 63 63 63
12 4
  0.4444
27 9
f. Are gender and wearing a seatbelt independent? INDEPENDENT
4 16
P(M)?P(M|SB)
?
9 36
0.4444  .44444
2.
Real estate ads suggest that 64% of homes for sale have attached garages, 21% have swimming pools,
and 17% have both features. What is the probability that a randomly selected home for sale has:
a. A pool or a garage? 0.47+0.17+.04=
0.68
Pool
Garages
b. Neither a pool nor a garage? 1- 0.68 = 0.32
0.47
0.17
0.04
P( Pool ' G ')  1  P( Pool  G )
0.32
c. A pool but no garage?
P( Pool  G ')  0.04
d. Are having a pool and having a garage independent?
P(Pool)= 0.21
P(Pool|Garage)= 0.17 Not =
NO
1
II. Tree Diagrams
Tree diagrams can be helpful when finding the probability of multiple events using conditional probability.
3.
The diagram shows the probability that a day will begin clear or cloudy, and then the probability of rain
on days that will begin clear and cloudy. The path containing clear and rain represents days that you
believe will start out clear and then it actually rains.
a. Find the probability of it being a cloudy day.
0.72
b. Find the probability the day starting clear AND then raining.
P(Clear  Rain)  (0.28)(0.04)  0.0112
c. Find the probability of rain, given that the day began cloudy.
P( Rain | Cloudy ) 
P(rain  cloudy ) (0.72)(0.31)

 0.31
P(cloudy )
0.72
d. Find the probability that it will rain given it began clear.
P( Rain | Clear ) 
P(rain  clear ) (0.28)(0.04)

0.04
P(clear )
0.28
e. Find the probability that it will rain.
P( Rain)  P(Clear  Rain)  P(Cloudy  Rain)
= (0.28)(0.04) +(0.72)(0.31)
=0.2334
Properties of Tree Diagrams
1st branch: Simple
Probability
2nd branch: Conditional
Probability , so formula needed
Sum of probabilities of branches:
1
2
4. The tree diagram below shows the probability of school being cancelled when the temperature outside in
August is above and below 95°F.
0.7
a. Complete the missing probabilities on the diagram.
b. Find the probability the temperature is above 95°F and
there is no school.
0.72
Above
0.3
(0.72)(0.7)= 0.504
c. Find the probability that the temperature is below 95°F
and there is no school.
0.28
No School
School
0.1
No School
0.9
School
Below
(0.28)(0.1)= 0.028
d. Find the probability that there is no school.
P(above  no school) or P(below  no school)
0.504+ 0.028 = 0.532
e. Find the probability that there is school, given that the temperature is above 95°F.
P(school|above)=
P( school  above) (0.72)(0.3)

 0.3
P(above)
(0.72)
5. The ELISA test can help detect whether people have the HIV virus. As with any medical test, ELISA can
give false positive and false negative results. From experience, medical researchers estimate that the ELISA
test has a 0.2% false positive rate and a 0.1% false negative rate. Suppose the ELISA test is used in an area
where 5% of the population has HIV.
a. Draw a tree diagram to represent this situation.
0.999 Test +
0.05
0.95
HIV+
0.001
0.002
Test Test +
HIV-
0.998
Test -
b. Find the probability that a person does not have the virus and tests positive.
(False positive)  P( HIV  Test )
= (0.95)(0.002)=0.0019
c. Find the probability that a randomly selected person from this area would test positive.
(Positive)  P( HIV  Test )  P( HIV  Test )
= (0.05)(0.999) + (0.95)(0.002)=0.05185
d. Given that a randomly selected person has tested positive, find the probability that the person has HIV.
P(HIV+|Test+) 
P( HIV  Test ) (0.05)(0.999)

=0.9634
P(Test )
0.05185
3
III. Practice
6. The senior class is 55% female, and 32% are females who play a competitive sport. Find the probability that
a student plays a competitive sport, given that the student is female.
P( F )  0.55
P(CS|F)=
P(F  CS)=0.32
P(CS  F ) 0.32

 0.5818
P( F )
0.55
7. A softball game has an 80% chance of being cancelled for a light drizzle and a 30% chance of being
cancelled for a heavy fog when there is no drizzle. There is a 70% chance of heavy fog and a 30% chance of
light drizzle.
a. Make a tree diagram for this situation.
Cancelled
0.3
0.7
0.3
Heavy
Fog
Light
Drizzle
0.7
Not Cancelled
0.8
Cancelled
0.2
Not Cancelled
b. Find the probability that the game will be cancelled.
P(Cancelled)=P(Heavy Fog  Cancelled)+P(Light Drizzle  Cancelled)
= (0.7)(0.3)+(0.3)(0.8) = 0.21 + 0.24 = 0.45
c. Find the probability there will be a light drizzle and the game will not be cancelled.
P(Light Drizzle  Not Cancelled) = (0.30)(0.20) = 0.06
d. Find the probability that the game is cancelled, given that there is heavy fog.
P(cancelled|heavy fog) =
P(cancelled  heavy fog) (0.70)(0.30)

 0.30
P(heavy fog)
(0.70)
e. Find the probability that there was drizzle, given that the game was cancelled.
P(drizzle|cancelled) =
P(drizzle  cancel) (0.30)(0.80) 0.24


 0.5333
P(cancelled)
(0.45)
0.45
Part b
4
8. A medical test is designed to detect the presence of a particular autoimmune disease. Suppose 30% of the
population has this disease. The medical test give false positives 8% of the time, and it gives false negatives
2% of the time. Draw a tree diagram of this information and answer each of the following questions.
0.98
0.30
0.70
Disease
0.02
Test +
Test -
0.08
Test +
0.92
Test -
Disease’
a. What is the probability that a randomly selected person tests positive for this disease?
P(Test )  P( D  )  P( D '  )
= (0.30)(0.98) +(0.70)(0.08)
= 0.294 +0.056 = 0.35
b. What is the probability that a person tests negative, given that the person has the disease?
P(Test  | Disease) 
P(test-  Disease) (0.30)(0.02)

 0.02
P( Disease)
(0.30)
c. What is the probability that the person does not have the disease and tests positive?
P( Disease ' Test )  (0.70)(0.08)  0.056
d. What is the probability that a person actually has the disease, given that the test result is positive?
P( Disease | Test ) 
P(Disease  Test+) (0.30)(0.98)

 0.84
P(Test)
(0.35)
Part a
5
9. A football team has a 70% chance of winning when it doesn’t snow, but only a 40% chance of winning
when it does snow. There is a game on Friday, and there is a 35% chance of snow.
0.4
0.35
Snow
0.6
0.7
0.65
Win
Win’
Win
Snow’
0.3
Win’
a. Find the probability that the team wins the game on Friday.
P(Win)  P( Snow  Win)  P( Snow ' Win)
= (0.35)(0.4) + (0.65)(0.70)=0.595
b. If the team wins the game, what is the probability that it was snowing?
P(Snow|Win)=
P( snow  win) (0.35)(0.4)

 0.2353
P( win)
0.595
6