7. Conditional Probability, Independence, and Trees
- Page ID
- 23289
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Contents 7A:
- Conditional Probability, Example 1 (https://youtu.be/nlGibQ0FNok)
- Conditional Probability, Example 2 (https://youtu.be/CTEfY511NSE)
- Conditional Probability, Example 3 (https://youtu.be/AlT8lJdZJ3E)
- Conditional Probability, Example 4 (https://youtu.be/dIcwlUuLyyA)
- Independent Events, Example 1 (https://youtu.be/xv0LSmsQ7Nc)
- Independent Events, Example 2 (https://youtu.be/5mtEApAMd9Q)
- Independent Events, Example 3 (https://youtu.be/-t3X1w_hiVA)
Prework 7A:
- 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 \(Pr(FF|\text{ at least one }F)=\frac{C(5,2)}{C(5,2)+C(5,1)\cdot C(3,1)}\).
Contents 7B:
- Trees and probability, Example 1 (https://youtu.be/MElgs4cyr3o)
- Trees and probability, Example 2 (https://youtu.be/sRBImVlnruc)
- Trees and probability, Important Facts (https://youtu.be/nW7DN-WajCQ)
- Trees and probability, Example 3 (https://youtu.be/K_ioVFHU4JM)
- Trees and conditional probability, Example 1 (https://youtu.be/-P1kYrXUcZI)
- Trees and conditional probability, Example 2 (https://youtu.be/pM9s4Oc4NZ4)
Prework 7B:
- Box 1 has 4 red markers, 5 green markers, and 2 black markers. Box 2 has 3 red markers, 2 green markers, and 4 black markers. A box is chosen at random and then a marker is drawn from the box. What is the probability a green marker is chosen?
- You first randomly select a fair coin or weighted coin (the probability of a getting a tails on the weighted coin is .3). Then you flip the coin twice and record the results. What is the probability the coin was weighted, given that both flips are heads?
- A box contains six red balls and two white balls. Two of the 8 balls are randomly selected from the box, one after another and without replacement. What is the probability that the first ball selected is red, given that the second ball selected is white?
Solutions
The solutions are given below.