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6. Introduction to Probability with Equally Likely Outcomes

  • Page ID
    22298
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    The following topics are included in this series of eight videos.

    1. Intro to Probability 1
    2. Intro to Probability 2
    3. Equally Likely Outcomes, Example 2
    4. Equally Likely Outcomes, Example 3
    5. Equally Likely Outcomes, Example 4
    6. Equally Likely Outcomes, Example 5
    7. Equally Likely Outcomes, Example 6
    8. Equally Likely Outcomes, Example 7

     

     

     

     

     

     

     

     

    PREWORK:

    1. Two fair six-sided dice are rolled. What is the probability that the sum of the numbers is 8?

    2. 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?

    3. Sarah has three snowflake ornaments and six snowman ornaments for her tree. She randomly selects four ornaments. What is the probability that she gets two snowflake ornaments and two snowman ornaments?

    Google form

    Solutions:

    1. 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 \(Pr(\text{sum is 8})=\frac{5}{36}\).
    2. 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 \(Pr(\text{excellent service only})=\frac{125}{500}\).
    3. We use the formula: \(Pr(\text{2 snowflake and 2 snowman})=\frac{n(\text{2 snowflake and 2 snowman})}{n(S)}\). Since she has 9 ornaments total and is selecting 4 (where she can't pick the same ornament twice and the order doesn't matter), so \(n(S)=C(9,4)\). For the numerator, she needs 2 snowflake ornaments, so that's \(C(3,2)\) and 2 snowman ornaments, so that's \(C(6,2)\). Since she needs both types, we multiply these together, hence the answer is \(Pr(\text{2 snowflake and 2 snowman})=\frac{C(3,2)\cdot C(6,2)}{C(9,4)}\).

    6. Introduction to Probability with Equally Likely Outcomes is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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