9: 8. Conditional Probability and Independence
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The following topics are included in this series of seven videos.
- Conditional Probability, Example 1
- Conditional Probability, Example 2
- Conditional Probability, Example 3
- Conditional Probability, Example 4
- Independent Events, Example 1
- Independent Events, Example 2
- Independent Events, Example 3
Prework:
- Suppose Pr(A)=0.4, Pr(B)=0.6, and Pr(A∩B)=0.15. Compute Pr(A|B) and Pr(A|Bc).
- Suppose that Pr(A)=0.3, Pr(B)=0.5, Pr(A∩Bc)=0.16. Are A and B independent? Explain.
- Suppose A and B are independent events, with Pr(A)=.4 and Pr(B)=.7. Compute Pr(Ac|B) and Pr(A|Bc).
- There 5 first-years and 3 sophomores in a classroom. Two are selected at random. What is the probability that both are first-years given that at least one is a first-year?
Solutions:
- Using a Venn diagram, we can determine that Pr(A∩B)=0.15, Pr(A∩Bc)=0.25, Pr(Ac∩B)=0.45, and Pr(Ac∩Bc)=0.15. Then Pr(A|B)=Pr(A∩B)Pr(B)=.15.6 and Pr(A|Bc)=Pr(A∩Bc)Pr(Bc)=.25.4.
- We need to check if Pr(A)⋅Pr(B)=Pr(A∩B). The left hand side of this equation just becomes 0.3⋅0.5=0.15. If we create and fill in a Venn diagram, we determine that Pr(A∩B)=0.14. Since 0.15≠0.14, A and B are NOT independent.
- Since A and B are independent events, Pr(A∩B)=Pr(A)⋅Pr(B)=0.4⋅0.7=0.28. We can use this information to fill out a Venn diagram and determine that Pr(Ac|B)=Pr(Ac∩B)Pr(B)=.42.7=0.6 and Pr(A|Bc)=Pr(A∩Bc)Pr(Bc)=.12.3=0.4. Another way to do this problem is to realize that since A and B are independent, so are Ac and B, and A and Bc (and, though it isn't relevant for this problem, Ac and Bc). Therefore, for the first part of the problem, we can rewrite the numerator as Pr(Ac|B)=Pr(Ac∩B)Pr(B)=Pr(Ac)⋅Pr(B)Pr(B)=Pr(Ac)=1−Pr(A)=.6. We can do the same thing for the second part of the problem: Pr(A|Bc)=Pr(A∩Bc)Pr(Bc)=Pr(A)⋅Pr(Bc)Pr(Bc)=Pr(A)=0.4.
- We use the conditional probability formula: Pr(FF| at least one F)=Pr(FF∩ at least one F)Pr( at least one F)=Pr(FF)Pr(FF or FS). The numerator is then Pr(FF)=C(5,2)C(8,2) and the denominator is Pr(FForFS)=C(5,2)+C(5,1)⋅C(3,1)C(8,2). Combining these, we get that Pr(FF| at least one F)=C(5,2)C(5,2)+C(5,1)⋅C(3,1).