Bayes' rule - Probability

Bayes' rule


• Multiplication Rule: P(AandB)=P(B)P(AB)P(A \;and\;B)=P(B) \cdot P(A|B)
• Conditional Probability: P(BA)P(B|A) == P(AandB)P(A)\frac{P(A \;and\; B)}{P(A)}
• Law of Total Probability: P(A)=P(B1)P(AB1)+P(B2)P(AB2)++P(Bn)P(ABn)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(BA)P(B|A) == P(AandB)P(A)=P(B)P(AB)P(A)\frac{P(A \;and\; B)}{P(A)}= \frac{P(B) \cdot P(A|B)}{P(A)}
=P(B)P(AB)P(B1)P(AB1)+P(B2)P(AB2)++P(Bn)P(ABn)=\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)}
  • 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.
  • 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.
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Bayes' rule

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