# 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 $$Pr(A\cap B)=0.15$$. 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?

#### Solutions:

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.