Conditional Probability
When dealing with two events (usually called A and B), sometimes the events are so related to each other, that the probability of one depends on whether the other event has occurred. When we talk about probabilities based on the fact that something else has already happened we call this conditional probability.

What changes when dealing with conditional probability is that we know for certain that something else has already happened. This means that in our definition of probability that says

where the “total number of ways” is based on the fact that we know something else has already occurred.

There are two ways to approach conditional probability and depends on the type of problem that you are given. In a situation where you are given percentages and probabilities (usually but not always in a table format) we make use of the conditional probability rule (shown below). In situations where you are trying to compute probabilities on your own (instead of them being given to you) most of the time it is easier to not to use the formula.

Conditional Probability Rule:

Consider events A and B.

The line between A and B is read “given”. So translated, this reads, "the probability of A given that B has happened." The event on the right side of the line is the event that has already happened.

What The Rule Means:

This rule is applied when you have two events and you already know the outcome of one of the events. In doing the computations, you will need to be able to find the probability of A and B, that is, P(AB). Problems of this type make use of the multiplication rule. If you need help with the multiplication rule or understanding what type of problems make use of the rule, review the lesson on the Multiplication Rule.

Let’s look at several examples that do and do not make use of the rule.

Let's Practice:

1. A survey of 500 adults asked about college expenses. The survey asked questions about whether or not the person had a child in college and about the cost of attending college. Results are shown in the table below.

 Cost Too Much Cost Just Right Cost Too Low Child in College 0.30 0.13 0.01 Child not in College 0.20 0.25 0.11

Suppose one person is chosen at random. Given that the person has a child in college, what is the probability that he or she ranks the cost of attending college as “cost too much”?

P(cost too much  child in college) or
P(cost too much given that there is a child in college)

According to the conditional probability rule:

P(cost too much child in college) =

P(cost too muchchild in college) can be found from the table as 0.30

P(child in college) can be found by adding 0.30 + 0.13 + 0.01 = 0.44

Substituting these values into the equation for conditional probability we get

=

In the previous example notice that the denominator of the probability fraction dealt with having a child in college. This was our given information. In other words, we already KNOW that the person has a child in college. So we aren’t talking about all 500 people in the sample, we’re only talking about the ones that have a child in college. In this problem we don’t know the actual numbers, but we do know percentages.

Just keep in mind that what conditional probability does for us is to adjust the denominator of our probability fraction to account for what we KNOW has already occurred.

Now let’s look at an example where probabilities are not already given and we have to come up with our own as we work through the problem.

1. Suppose you draw two cards from a standard deck without replacement. Given that the first card is an ace, what is the probability that the second card is a queen?

Let’s take a look at this problem without making use of the formula. Consider that we KNOW that an ace has already been pulled from the deck. This means there are now 3 aces in the deck of 51 cards that are left. Now consider the probability of drawing a queen from that deck of 51. There are still four queens in the remaining deck of 51. This gives the probability of 4/51.

The problem could be solved using the conditional probability rule as shown below. However, look at how much work is needed to use the formula rather than simply thinking through what the problem says.

P(ace given queen) or P(ace queen)

According to the conditional probability rule

P(ace queen) =

P(acequeen) is found using the multiplication rule. Since the problem states that there is no replacement we can find

P(acequeen) as (4/52)(4/51) = 4/663

We know that

P(queen) = 4/52

Substituting values into the conditional probability formula we get

So how will you make the determination of how to solve a conditional probability problem? As a general rule, problems where information is given in a table are best solved by using the formula. Problems where you have to figure out probabilities as you work through the problem are best solved by adjusting for the information you already know.

Example Group #1
The first three problems use the table given in the first example of Let's Practice.

 Cost Too Much Cost Just Right Cost Too Low Child in College 0.30 0.13 0.01 Child not in College 0.20 0.25 0.11

Suppose one person is chosen at random.
 Find the probability that a person thinks college costs are just right given that they have a child in college. What is your answer?
 Find the probability that the person thinks college costs are too low given that they do not have a child in college. What is your answer?
 Find the probability that the person does not have a child in college thinks college costs are too low. What is your answer?

Examples
 In the United States, 55% of children get an allowance and 41% of children get an allowance and do household chores. What is the probability that a child does household chores given that the child gets an allowance? What is your answer?
 At a large high school, the probability that a student takes Japanese and Spanish is 0.08. The probability that a student takes Spanish is 0.45. What is the probability that a student taking Japanese given that the student is taking Spanish? What is your answer?

Example Group #3
The final two problems involve dice.
 Consider rolling two dice. What is the probability that the sum is odd given that one die shows a 3? What is your answer?
 Consider rolling two dice. What is the probability that the sum is even given that one die shows a 3? What is your answer?

S Taylor

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