7: 6. Introduction to Probability with Equally Likely Outcomes
 Page ID
 25692
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The following topics are included in this series of eight videos.
 Intro to Probability 1
 Intro to Probability 2
 Equally Likely Outcomes, Example 2
 Equally Likely Outcomes, Example 3
 Equally Likely Outcomes, Example 4
 Equally Likely Outcomes, Example 5
 Equally Likely Outcomes, Example 6
 Equally Likely Outcomes, Example 7
PREWORK:

Two fair sixsided dice are rolled. What is the probability that the sum of the numbers is 8?

In a survey of 500 patrons of a local restaurant, we find the following: 300 say the food is excellent, 275 say the service is excellent, and 150 say both the service and food are excellent. What is the probability that one of these patrons chosen at random says that only the service is excellent?

A student is planning to do 4 errands over break: visit her grandmother, see a movie, go the the bank and shop at the mall. If she randomly orders the errands, what is the probability that she visits her grandmother before she goes to the mall and the bank?
Solutions:
 We use the formula for probability with equally likely outcomes: \(Pr(\text{sum is 8})=\frac{n(\text{sum is 8})}{n(S)}\). By the multiplication principle there \(6\cdot 6=36\) possible outcomes when rolling two dice, so \(n(S)=36\). For the numerator, we simply count: the set of possible outcomes which give a sum of 8 is \(\{(6,2), (5,3), (4,4), (3,5), (2,6)\}\). There are five of these, so the numerator is \(5\). Hence \(Pr(\text{sum is 8})=\frac{5}{36}\).
 First make a Venn diagram with 2 sets (one for those who say the service was excellent and one for those who say the food was excellent). Once we fill in the Venn diagram, we notice that exactly 125 said that only the service was excellent. Therefore \(Pr(\text{excellent service only})=\frac{125}{500}\).
 We let \(E\) be the event where she visits her grandmother before going to the mall and bank. Then \(Pr(E)=\frac{n(E)}{n(S)}\). Since she is doing \(4\) errands, there are \(4!=24\) possible orders in which she can do the errands (so the denominator is 24). If she goes to visit her grandmother as the first errand, then she can perform the remaining 3 errands in any order and still visit her grandmother prior to going to the mall and bank. There are \(3!=6\) in which this could happen. She could also visit her grandmother second, immediately after going to see a movie. This would give 2 ways for her to see her grandmother prior to the bank and mall. Therefore \(n(E)=8\), and the answer is \(Pr(E)=\frac{8}{24}\).