Conditional Probability — Part 3

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Name:
Date: _______________
Algebra 2
Conditional Probability 3
Conditional Probability — Part 3
Objective: You will calculate the probability of a top-level event
in a probability tree, given that a bottom-level event occurs.
So far in our work with conditional probability, we’ve learned
methods for finding “and” probabilities, such as
P(A and C) = P(A) · P(C | A),
as well as methods for finding probabilities of second-level events,
such as
P(C) = P(A and C) + P(B and C) = P(A) · P(C | A) + P(B) · P(C | B).
But here’s something we don’t know how to do yet. Suppose C happens. Then what’s the
probability that C happened following A, versus happened following B? For example, in
problem 6 about disease testing: suppose someone tests positive; what’s the probability that they
actually have the disease? Obviously that’s an important question, and the answer is often
surprising and not-well-understood.
Example: Disease Testing
Here’s a recap of the disease testing problem, including a
chance to check the answers to parts of question 6. The tree at
the right shows all the probabilities (check answer to 6a).
Let’s check probabilities associated with having a positive test:
6b. P(disease and positive) = 0.03 · 0.98 = 0.0294
6c. P(no-disease and positive) = 0.97 · 0.05 = 0.0485
6d. P(positive) = 0.0294 + 0.0485 = 0.0779
Remarkably, false positive results happen more often than accurate positive tests. (This happens
on many real tests for diseases, and many people are unaware of this fact.)
Now here’s the new type of question you will learn to answer today: What is the probability that
someone who tests positive actually has the disease? To get the answer, you divide the
probability of an accurate positive result by the probability of any positive result:
P(disease | positive ) 
P(disease and positive ) 0.0294

 0.377
P(positive )
0.0779
So in this problem there’s about a 37.7% chance that someone who tests positive actually has the
disease.
Here’s the intuition behind the calculation. Think about the probabilities as percentages and
round them off. Per 100 people, there are 3 accurate positives and 5 false positives, making
8 total positives. So on a positive test, there’s a 3/8 chance it’s accurate and a 5/8 chance it’s
false.
Due to this probability situation, the initial response to a positive medical test is often to follow
up with a more accurate test or other medical analysis, before treating the disease.
Name:
Date: _______________
Algebra 2
Conditional Probability 3
You can do the same kind of calculations involving negative results:
6e. accurate negative result: 0.97 · 0.95 = 0.9215
6f. false negative result:
0.03 · 0.02 = 0.0006
6g. all negative results:
0.9215 + 0.006 = 0.9221
Note that false negatives are rare (fortunately, because they could leave the patient with an
undetected, untreated disease — did you say that in 13k?).
Now try answering this question. Hint: You just have to divide two of the probabilities shown above.
You try it
For the disease testing in problem 13, suppose someone gets a negative test result. What is the
probability that the person does not have the disease?
General Result
Here is the general version of how to do problems like the preceding
example. The goal is to find the probability of a top-level event given
that a bottom-level event happens.
Here is how to find P(A | C), the probability of A given that
C happens:
P(A | C) 
P(A and C)
.
P(C)
Pay attention to the distinction between P(C | A) and P(A | C).
P(C | A) is just one of the probabilities in the tree (on the C branch underneath A).
P(A | C) has to be calculated using the formula shown above.
Similar calculations are available for all the other probabilities of top-level events given
bottom-level events. Here’s one of the others:
P ( B | D) 
P(B and D)
.
P(D)
Name:
Date: _______________
Algebra 2
Conditional Probability 3
Problems
7. Suppose that there’s another disease test that has better accuracy. As before, suppose that 3%
have the disease and 97% do not. When someone with the disease is tested, there’s a 99%
chance that the test result is positive. When someone not having the disease is tested, there’s
a 99.2% chance that the test result is negative.
a. Label each branch of this tree diagram with a probability.
b. What is the probability that the test gives an accurate
positive result?
c. What is the probability that the test result is positive?
d. What is the probability that someone who tests positive actually has the disease?
e. What is the probability that someone who tests negative actually has the disease?
(Hint: You need to go through steps like in parts b, c, and d again, but with the negative
probabilities.)
Name:
Date: _______________
Algebra 2
Conditional Probability 3
8. Answer the following questions about this probability tree:
a. Find P(C | A).
b. Find P(A and C).
c. Find P(C).
d. Find P(A | C). That is, supposing that C happens, what is the probability of A ?
(Hints: The answer is not 0.2. It might help to think of A as “disease” and C as “positive”
and follow the method of the example on page 1.)
e. Find P(D | B).
f. Find P(B | D).
Name:
Date: _______________
Algebra 2
Conditional Probability 3
9. During a flu epidemic, 35% of a school’s students have the flu. Of those with the flu,
90% have high temperatures. However, a high temperature is also possible for people
without the flu. The school nurse estimates that 12% of those without the flu have high
temperatures too.
a. Draw a tree diagram to represent this situation and label it with probabilities.
(Hint: Think of “high temperature” as being a test for the flu.)
b. What percent of students have a high temperature?
c. If a student has a high temperature, what is the probability that the student has the flu?
Name:
Date: _______________
Algebra 2
Conditional Probability 3
10. Here’s the three-jar situation from problem 12 of the last
assignment. A die is rolled.
If the roll is 1, 2, or 3, a ball is drawn from jar A.
If the roll is 4 or 5, a ball is drawn from jar B.
If the roll is 6, a ball is drawn from jar C.
a. When the die is rolled and a ball is drawn, what is the
probability that it is a white ball from Jar A?
b. When the die is rolled and a ball is drawn, what is the probability that it is a white ball
(from any of the jars)?
c. Suppose that a white ball is drawn. What is the probability that it came from Jar A?
d. Suppose that a black ball is drawn. What is the probability that it came from Jar C?
e. Suppose that a black ball is drawn. What is the probability that it did not come from Jar C?
(Hint: Use answer to d.)
Name:
Date: _______________
Algebra 2
Conditional Probability 3
11. Machine A produces 60% of the ball bearings manufactured by a factory and Machine B
produces the rest. Five percent of Machine A’s bearings fail (are produced incorrectly) and
two percent of Machine B’s bearings fail.
a. Make a probability tree representing this situation.
b. What is the probability that a bearing fails?
c. If a bearing fails, what is the probability that it was produced by Machine A?
12. Five percent of the welds on an automobile assembly line are defective. The defective welds
are found using an X-ray machine. The machine correctly rejects 92% of the defective welds
and correctly accepts all of the good welds.
a. Label this tree diagram with probabilities.
b. What is the probability that an accepted weld is actually defective?
Name:
Date: _______________
Algebra 2
Conditional Probability 3
Answers
7. a. probabilities should be: 0.03, 0.97; 0.99, 0.01; 0.008, 0.992
b. 0.03 · 0.99 = 0.0297
c. 0.03 · 0.99 + 0.97 · 0.008 = 0.03746
d. Divide answers to b and c, get 0.7928
e.
0.03  0.01
 0.00031
0.03  0.01  0.97  0.992
8. a. 0.9
b. 0.2 · 0.9 = 0.18
c. 0.2 · 0.9 + 0.8 · 0.7 = 0.18 + 0.56 = 0.74
d.
0.18
 0.2432
0.74
e. 0.3
f.
P( B and D)
0.8  0.3
0.24


 0.9231
P( D)
0.2  0.1  0.8  0.3 0.26
9. a. probabilities should be: 0.35, 0.65; 0.9, 0.1, 0.12, 0.88
b. 0.35 · 0.9 + 0.65 · 0.12 = 0.393, so 39.3% of students have a high temperature.
c.
P( flu and hightemp)
0.35  0.9
0.315


 0.802 , or about 80%.
P(hightemp)
0.35  0.9  0.65  0.12 0.393
10. a.
3
6
 23 
b.
3
6
 23  62  24  16  14  13  16  241 
c.
1
3
13
24
1
3
13
24
 138
d. Calculate P(Jar C and black) divided by P(black from any jar).
e. 1  113  118 .
11. a. should have probabilities 0.6, 0.4; 0.95, 0.05, 0.98, 0.02
b. 0.6 · 0.05 + 0.4 · 0.02 = 0.03 + 0.008 = 0.038
c.
0.03
 0.7895
0.038
12. a. should have probabilities 0.95, 0.05; 1, 0, 0.08, 0.92
b.
0.05  0.08
0.004

 0.0042
0.95  1  0.05  0.08 0.954
3
24
11
24
 113 .
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