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Bayes' rule

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Intros
Lessons
  1. Deriving Bayes' rule
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Examples
Lessons
  1. Bayes' Rule
    I am going to ask my boss to be my reference after applying to another job. If she gives me a good recommendation there is a 0.75 probability that I will get the job. While if she gives me a bad recommendation there is only a 0.25 probability that I will get the job. There is a 60% chance she will give me a good reference and a 40% chance she will give me a bad reference.
    1. What is the probability that I will get the job?
    2. Given that I got the job what is the probability that she gave me a good reference
  2. I have 3 bags that each contains 5 marbles.

    Bag A:

    Bag B:

    Bag C:

    2 Green

    4 Green

    5 Green

    3 Red

    1 Red

    0 Red


    I roll a fair die to decide which bag I will draw from. If I roll a 1,2,3 I will draw a marble from Bag A. If I roll a 4,5 I will draw from Bag B. And if I roll a 6, then I will draw a marble from bag C.
    1. What is the probability that I draw a red marble?
    2. Suppose that I drew a green marble from a bag. What is the probability that I rolled a 6?
  3. False Positives
    A blood test is 95% effective when diagnosing a diseased person. However this blood test also incorrectly diagnoses a healthy person 5% of the time. If 0.1% of the population actually has this disease, then what is the probability that a person has the disease given that they tested positive?

    I like this, gives you hope if you get diagnosed for a serious disease
    Make note of how many healthy people and how many diseased people get diagnosed. Also make a tree diagram illustrating this.
    Topic Notes
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    Recall:
    • Multiplication Rule: P(A  and  B)=P(B)⋅P(A∣B)P(A \;and\;B)=P(B) \cdot P(A|B)
    • Conditional Probability: P(B∣A)P(B|A) == P(A  and  B)P(A)\frac{P(A \;and\; B)}{P(A)}
    • Law of Total Probability: P(A)=P(B1)P(A∣B1)+P(B2)P(A∣B2)+⋯+P(Bn)P(A∣Bn)P(A)=P(B_1)P(A|B_1)+P(B_2)P(A|B_2)+ \cdots+P(B_n)P(A|B_n)

    Combining all these equations we get Bayes' Rule:
    P(B∣A)P(B|A) == P(A  and  B)P(A)=P(B)⋅P(A∣B)P(A)\frac{P(A \;and\; B)}{P(A)}= \frac{P(B) \cdot P(A|B)}{P(A)}
    =P(B)⋅P(A∣B)P(B1)P(A∣B1)+P(B2)P(A∣B2)+⋯+P(Bn)P(A∣Bn)=\frac{P(B) \cdot P(A|B)}{P(B_1)P(A|B_1)+P(B_2)P(A|B_2)+ \cdots+P(B_n)P(A|B_n)}