Math 207 –
Conditional Probability (General Multiplication Rule; Bayes’
Rule) Examples
Important Note: Although these problems illustrate the Law of Total Probability and Bayes’ Rule, you need not memorize the
formula for the rule—in fact, I encourage you to solve each problem
individually via a tree diagram.
- A bag contains 20 jelly beans: 5
black and 15 red beans. Two jelly beans are drawn at random and without
replacement from the bag. Consider the events: A = {2 black jelly beans},
B = {exactly 1 black jelly bean}, and C = {0 black jelly beans}.
- Use counting methods
(combinations—order unimportant) to find P(A),
P(B), and P(C).
- Now use the general multiplication
rule to determine P(A), P(B), and P(C). Do your
answers agree? (They should.)
- Draw a complete tree diagram to
illustrate your answer to part b.
(Note: In the future, it’s fine for you to solve problems using a tree
diagram as your argument—as long as
the diagram is well-labeled, complete, and correct.)
- The probability of a certain
disease is 0.01. A diagnostic test for the disease is developed. The test
correctly diagnoses an infected person with probability 0.95. However, the
test incorrectly diagnoses an uninfected patient with probability 0.06. If
the test diagnoses a patient as having the disease, what is the
conditional probability that the person really has the disease? Does this
probability seem low? Why do you think that is?
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Law of
Total Probablity
Suppose the events
are mutually exclusive and
(that is, the events “exhaust” the sample
space). [Note: In the previous example, the mutually exclusive and exhaustive
events were {disease} and {no disease}.]
Then for any event B,
, where first we
applied Axiom 3 for the probability of mutually exclusive events and then
applied the multiplicative rule.
That is,
This is called the
Law of Total Probability and it leads to another formula called Bayes’ Rule. You need not memorize these formulas, but
you must understand the concept and be able to apply the general concept to a
new problem—it’s fine to use a complete, well-documented tree diagram as your
solution method.
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- About 99% of babies born in the United States
survive. Furthermore, Caesarian section is used in about 20% of births.
Given a Caesarian section is used, about 96% of babies survive (note that
while the previous two numbers are fairly accurate, this number is
actually made up). Now, given that a Caesarian section is not used, about what percent of
babies survive? (Note: Now you’re looking for a probability within the
tree diagram.)