9: 8. Conditional Probability and Independence
- Page ID
- 25694
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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 \(Pr(A\cap B)=0.15\). Compute \(Pr(A|B)\) and \(Pr(A|B^c)\).
- Suppose that \(Pr(A) = 0.3\), \(Pr(B) = 0.5\), \(Pr(A\cap B^c) = 0.16\). Are \(A\) and \(B\) independent? Explain.
- Suppose \(A\) and \(B\) are independent events, with \(Pr(A) = .4\) and \(Pr(B) = .7\). Compute \(Pr(A^c|B)\) and \(Pr(A|B^c )\).
- 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 \(Pr(A\cap B)=0.15\), \(Pr(A\cap B^c)=0.25\), \(Pr(A^c\cap B)=0.45\), and \(Pr(A^c\cap B^c)=0.15\). Then \(Pr(A|B)=\frac{Pr(A\cap B)}{Pr(B)}=\frac{.15}{.6}\) and \(Pr(A|B^c)=\frac{Pr(A\cap B^c)}{Pr(B^c)}=\frac{.25}{.4}\).
- 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.
- 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: \(Pr(A|B^c)=\frac{Pr(A\cap B^c)}{Pr(B^c)}=\frac{Pr(A)\cdot Pr(B^c)}{Pr(B^c)}=Pr(A)=0.4\).
- 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 \(Pr(FF|\text{ at least one }F)=\frac{C(5,2)}{C(5,2)+C(5,1)\cdot C(3,1)}\).