AP Statistics Section 6.3C More Conditional Probability

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Transcript AP Statistics Section 6.3C More Conditional Probability

AP Statistics Section 6.3C
More Conditional Probability
Example: Assume that 1.3% of AHS students smoke
on a daily basis. A breath test is designed to
determine if someone has smoked in the last week.
Assume the test will come back positive 96% of the
time if someone has smoked in the last week but will
also come back positive 7% of the time if someone
has not smoked in the last week (a false positive).
Find the probability that a randomly selected student
at AHS has smoked in the last week if the test comes
back positive? We will use a tree diagram to organize
our thinking.
Assume that 1.3% of AHS students smoke on a daily basis. A breath test is designed to
determine if someone has smoked in the last week. Assume the test will come back
positive 96% of the time if someone has smoked in the last week but will also come
back positive 7% of the time if someone has not smoked in the last week (a false
positive). Find the probability that a randomly selected student at AHS has smoked in
the last week if the test comes back positive? We will use a tree diagram to organize
our thinking.
positive
.96
.013
AHS
Students
.987
smoke
negative
.04
.07
positive
don' t smoke
.93
negative
Assume that 1.3% of AHS students smoke on a daily basis. A breath test is designed to
determine if someone has smoked in the last week. Assume the test will come back
positive 96% of the time if someone has smoked in the last week but will also come
back positive 7% of the time if someone has not smoked in the last week (a false
positive). Find the probability that a randomly selected student at AHS has smoked in
the last week if the test comes back positive? We will use a tree diagram to organize
our thinking.
.96
.013
AHS
Students
.987
smoke
.04
.07
positive
 .01248
negative
 .00052
positive
 .06909
negative
 .91791
don' t smoke
.93
P( smoked / positve) 
smoked  positive
.01248


.
01248

.
06909
positive
.153
Example: Online chat rooms are dominated by
the young. Teens are the biggest users. If we
look only at adult internet users (aged 18 and
older), 47% of the 18 to 29 age group chat, as do
21% of those aged 30 to 49 and just 7% of those
50 and over. To learn what percent of all adult
internet users participate in chat rooms, we also
need the age breakdown of users: 29% of adult
users are aged 18 to 29, 47% are aged 30 to 49
and 24% are 50 and over. We will use a tree
diagram to organize our thinking.
Chat .47
No .53
18  29
.29
.1537
Chat .21
.0987
No .79
.3713
30  49
.47
over 50
.24
.1363
Chat .07
No .93
.0168
.2232
Are the events “18-29 year old internet user”
and “adult chat room user” independent”?
P(18  29  chat ) .1363
P(18  29 / chat ) 

 .5413
.2518
P(chat )
P(18  29)  .29
P(18  29)  P(18  29 / chat )
.29  .5413
NO