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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









    1. Suppose \(Pr(A)=0.4\), \(Pr(B)=0.6\), and \(Pr(A\cap B)=0.15\). Compute \(Pr(A|B)\) and \(Pr(A|B^c)\).
    2. Suppose that \(Pr(A) = 0.3\), \(Pr(B) = 0.5\), \(Pr(A\cap B^c) = 0.16\). Are \(A\) and \(B\) independent? Explain.
    3. Suppose \(A\) and \(B\) are independent events, with \(Pr(A) = .4\) and \(Pr(B) = .7\). Compute \(Pr(A^c|B)\) and \(Pr(A|B^c )\).
    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?


    1. Using a Venn diagram, we can determine that \(Pr(A\cap B)=0.15\), \(Pr(A\cap B^c)=0.25\), \(Pr(A^c\cap B)=0.45\), and \(Pr(A^c\cap B^c)=0.15\). Then \(Pr(A|B)=\frac{Pr(A\cap B)}{Pr(B)}=\frac{.15}{.6}\) and \(Pr(A|B^c)=\frac{Pr(A\cap B^c)}{Pr(B^c)}=\frac{.25}{.4}\).
    2. We need to check if \(Pr(A)\cdot Pr(B)=Pr(A\cap B).\) The left hand side of this equation just becomes \(0.3\cdot 0.5=0.15\). If we create and fill in a Venn diagram, we determine that \(Pr(A\cap B)=0.14\). Since \(0.15\neq 0.14\), \(A\) and \(B\) are NOT independent.
    3. Since \(A\) and \(B\) are independent events, \(Pr(A\cap B)=Pr(A)\cdot Pr(B)=0.4\cdot 0.7=0.28\). We can use this information to fill out a Venn diagram and determine that \(Pr(A^c|B)=\frac{Pr(A^c\cap B)}{Pr(B)}=\frac{.42}{.7}=0.6\) and \(Pr(A|B^c)=\frac{Pr(A\cap B^c)}{Pr(B^c)}=\frac{.12}{.3}=0.4\). Another way to do this problem is to realize that since \(A\) and \(B\) are independent, so are \(A^c\) and \(B\), and \(A\) and \(B^c\) (and, though it isn't relevant for this problem, \(A^c\) and \(B^c\)). Therefore, for the first part of the problem, we can rewrite the numerator as \(Pr(A^c|B)=\frac{Pr(A^c\cap B)}{Pr(B)}=\frac{Pr(A^c)\cdot Pr(B)}{Pr(B)}=Pr(A^c)=1-Pr(A)=.6\). We can do the same thing for the second part of the problem: \(Pr(A|B^c)=\frac{Pr(A\cap B^c)}{Pr(B^c)}=\frac{Pr(A)\cdot Pr(B^c)}{Pr(B^c)}=Pr(A)=0.4\).
    4. We use the conditional probability formula: \(Pr(FF|\text{ at least one }F)=\frac{Pr(FF\cap \text{ at least one }F)}{Pr(\text{ at least one }F)}=\frac{Pr(FF)}{Pr(FF\text{ or }FS)}\). The numerator is then \(Pr(FF)=\frac{C(5,2)}{C(8,2)}\) and the denominator is \(Pr(FF or FS)=\frac{C(5,2)+C(5,1)\cdot C(3,1)}{C(8,2)}\). Combining these, we get that \(Pr(FF|\text{ at least one }F)=\frac{C(5,2)}{C(5,2)+C(5,1)\cdot C(3,1)}\).


      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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