Solving a Word Problem Involving the Probability of a Union

Solving a Word Problem Involving the Probability of a Union

Step 1: Identify the two events relevant to the problem.

Step 2: Determine the probability of each event occurring alone.

Step 3: Calculate the probability of the intersection of the two events.

Step 4: Calculate the probability of the union by adding the probabilities of the two events occurring alone and subtracting the probability of the intersection of the two events.

Solving a Word Problem Involving the Probability of a Union Vocabulary and Formulas

Event: The outcome(s) of a random experiment

Random Experiment: A repeatable procedure that produces outcomes that cannot be predicted in the short-run, but exhibit patterns in its long-run behavior

Union: The union of two events is the set of all outcomes that can occur in either event. It is the set of outcomes in the first event or the second event.

Intersection: The intersection of two events is the set of all outcomes the two events have in common. It is the set of outcomes in both the first event and the second event.

Disjoint: Two events are disjoint, or mutually exclusive, if they have no outcomes in common. The probability of the intersection of disjoint events is always zero.

Now, we are ready to look at two examples of solving a word problem involving the probability of a union, one where the two events have outcomes in common, and on where the two events are disjoint.

Solving a Word Problem Involving the Probability of a Union: Two Events with Outcomes in Common Example

Consider rolling a fair six-sided die. In a single roll, what is the probability of rolling a 3 or an odd number?

Step 1: First, we identify the two events relevant to the problem.

First Event: Rolling a 3 when a fair six-sided die is rolled

Second Event: Rolling an odd number when a fair six-sided die is rolled.

Step 2: Determine the probability of each event occurring alone.

P(3) = {eq}\frac{1}{6} {/eq} since there is one way to roll a 3 and six total outcomes, 1, 2, 3, 4, 5, or 6.

P(odd number) = {eq}\frac{3}{6}=\frac{1}{2} {/eq} since there are three ways to roll an odd number, 1, 3, or 5 and six total outcomes.

Step 3: Calculate the probability of the intersection of the two events.

P(3 and odd number) = ? since the two events have one outcome in common, 3.

Step 4: Calculate the probability of the union by adding the probabilities of the two events occurring alone and subtracting the probability of the intersection of the two events.

P(3 or odd number) = P(3) + P(odd number) - P(3 and odd number)

P(3 or odd number) = {eq}\frac{1}{6}+\frac{1}{2}-\frac{1}{6}=\frac{1}{2} {/eq} or 0.5

So, the probability of rolling a 3 or an odd number is 0.5.

Solving a Word Problem Involving the Probability of a Union: Two Disjoint Events Example

There is a bucket with 20 balls in it. Six of the balls are solid red, one of the balls is red with stripes, six of the balls are solid green, three balls are green with stripes, and four of the balls are solid yellow. After the balls are mixed, a child gets to reach in without looking to grab a ball to have as his own. What is the probability the child will grab a ball that is yellow or has stripes?

Step 1: Identify the two events relevant to the problem.

Event 1: The ball is yellow

Event 2: The ball has stripes

Step 2: Determine the probability of each event occurring alone.

P(yellow) = {eq}\frac{4}{20} {/eq}= 0.2 since there are four yellow balls and 20 total balls

P(stripes)= {eq}\frac{4}{20} {/eq} = 0.2 since there are four balls with stripes (one red with stripes and three green with stripes) and 20 total balls.

Step 3: Calculate the probability of the intersection of the two events.

There are no balls that are yellow with stripes. This means that the two events are disjoint and P(yellow and stripes) = 0.

Step 4: Calculate the probability of the union by adding the probabilities of the two events occurring alone and subtracting the probability of the intersection of the two events.

P(yellow or stripes) = P(yellow) + P (stripes) - P(yellow and stripes)

P(yellow or stripes) = 0.2 + 0.2 - 0 = 0.4

The probability the child will grab a ball that is yellow or has stripes is 0.4.