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Mathematics LibreTexts

9: 8. Conditional Probability and Independence

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The following topics are included in this series of seven videos.

  1. Conditional Probability, Example 1
  2. Conditional Probability, Example 2
  3. Conditional Probability, Example 3
  4. Conditional Probability, Example 4
  5. Independent Events, Example 1
  6. Independent Events, Example 2
  7. Independent Events, Example 3

 

 

 

 

 

 

 

Prework:

  1. Suppose Pr(A)=0.4, Pr(B)=0.6, and Pr(AB)=0.15. Compute Pr(A|B) and Pr(A|Bc).
  2. Suppose that Pr(A)=0.3, Pr(B)=0.5, Pr(ABc)=0.16. Are A and B independent? Explain.
  3. Suppose A and B are independent events, with Pr(A)=.4 and Pr(B)=.7. Compute Pr(Ac|B) and Pr(A|Bc).
  4. 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:

  1. Using a Venn diagram, we can determine that Pr(AB)=0.15Pr(ABc)=0.25Pr(AcB)=0.45, and Pr(AcBc)=0.15. Then Pr(A|B)=Pr(AB)Pr(B)=.15.6 and Pr(A|Bc)=Pr(ABc)Pr(Bc)=.25.4.
  2. We need to check if Pr(A)Pr(B)=Pr(AB). The left hand side of this equation just becomes 0.30.5=0.15. If we create and fill in a Venn diagram, we determine that Pr(AB)=0.14. Since 0.150.14, A and B are NOT independent.
  3. Since A and B are independent events, Pr(AB)=Pr(A)Pr(B)=0.40.7=0.28. We can use this information to fill out a Venn diagram and determine that Pr(Ac|B)=Pr(AcB)Pr(B)=.42.7=0.6 and Pr(A|Bc)=Pr(ABc)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(AcB)Pr(B)=Pr(Ac)Pr(B)Pr(B)=Pr(Ac)=1Pr(A)=.6. We can do the same thing for the second part of the problem: Pr(A|Bc)=Pr(ABc)Pr(Bc)=Pr(A)Pr(Bc)Pr(Bc)=Pr(A)=0.4.
  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).

 


    9: 8. Conditional Probability and Independence is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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