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Math 1313 Section 7. Conditional Probability
We’ll start this section with an example.
Example 1 : Two cards are drawn at random and without replacement from a well-shuffled deck of 52 playing cards. a. What is the probability that the first card drawn is an ace?
b. What is the probability that the second card drawn is an ace given that the first card drawn was not an ace?
c. What is the probability that the second card drawn is an ace given that the first card drawn was an ace?
In this experiment, we start with a sample space which contains 52 elements. We have two events, the first draw and the second draw. Since there is no replacement, the number in the sample space becomes smaller. In parts b and c of Example 1, we learn more about the experiment, which changes the sample space, which changes the probabilities.
These are examples of conditional probability.
We can demonstrate conditional probability using Venn diagrams.
Suppose we have an experiment with sample spaceS and supposeE and
F are events of the experiment. We can draw a Venn diagram of this
situation.
Now, suppose we know that eventE has occurred. This gives us this
picture:
So, the probability thatF occurs, given thatE has already occurred can be
expressed as ( ) ( ) ( ) ( ) ( | ). ( ) (^ ) ( ) ( )
n E F n E F n S P E F P F E n E n E P E n S
Conditional Probability of an Event
IfE andF are events of an experiment with P E ( ) ≠ 0 , then the conditional
probability that the eventF will occur given that the eventE has already
occurred is ( ) ( | ). ( )
P E F
P F E
P E
Example 4: A pair of fair dice is tossed and the number on the uppermost face is observed. What is the probability that the sum of the numbers falling uppermost is 6 if is it known that one of the numbers was 2?
Example 5 : A survey showed that 40% of all convenience store shoppers buy milk, 30% buy bread and 25% buy both milk and bread. a. If a randomly selected shopper buys milk what is the probability that s/he will also buy bread? b. What is the probability that a randomly selected shopper buys only bread?
Product Rule
Sometimes we know the conditional probability and are interested in finding. We can solve the conditional probability formula for
to get the product rule.
P E ( ∩ F
F )
P E ( ∩
P E F
P F E
P E
P E ( I F ) = P E ( ) * P F ( | E )
We will use tree diagrams for these kinds of problems to help organize the information we know. The first branch of the tree is the first trial and the second branch is the second. For the above formula, we could illustrate the product rule by:
Example 6 : An urn contains 5 blue marbles and 7 green marbles. a. Two marbles are drawn in succession and without replacement from the urn. What is the probability that both marbles are green?
b. Two marbles are drawn in succession without replacement from the urn. What is the probability that the second marble is green?
Example 9 : Mary K. Cosmetics estimates that 29% of the country has
s
Independent Events
wo eventsA andB are^ independent^ if the outcome of one event does
est for Independence of Two Events
wo eventsA andB are independent if and only if ).
seen its commercial and if a person sees its commercial, there is a 13% chance that the person will not buy its products. The company also claim that if a person does not see its commercial, they still have a 24% chance of buying the company’s products. What is the probability that a randomly selected person in the country will not buy its products?
T
not depend on the outcome of the other event.
T
T P A ( ∩ B ) = P A ( ) ⋅ P B (
This can be extended to any finite number of events.
Example 10 : Determine if the two stated events are independent. The experiment is drawing a card from a well-shuffled deck of 52 playing cards.
a. A = the event of drawing a face card
B = the event of drawing a heart
b. C = the event of drawing a heart
D = the event of drawing a club
c. E = the event of drawing a king
F = the event of drawing a red card
Note: It is possible for the events to overlap and still be independent. Many students think that “mutually exclusive events” and “independent events” mean the same thing. They do not.
Example 11 : Two eventsA andB are independent. P A ( ) =.43and
P B ( ) =.31. Find P A ( ∪ B ).
