a.) what is the probability that a female is ...

# Explanation: ## Step1: Recall probability formula $P(A)=\frac{n(A)}{n(S)}$, where $n(A)$ is the number of elements in event $A$ and $n(S)$ is the total number of elements in the sample - space. ## Step2: Calculate probability of choosing a female (a) The number of females $n(F)=2427$, and the total number of students $n(S)=3537$. So $P(F)=\frac{2427}{3537}\approx0.686$. ## Step3: Calculate probability of female or education - major (b) Use the formula $P(F\cup E)=P(F)+P(E)-P(F\cap E)$. $P(F)=\frac{2427}{3537}$, $P(E)=\frac{795}{3537}$, $P(F\cap E)=\frac{700}{3537}$. Then $P(F\cup E)=\frac{2427 + 795-700}{3537}=\frac{2522}{3537}\approx0.713$. ## Step4: Calculate probability of male and non - education major (c) The number of males who are non - education majors $n(M\cap\overline{E}) = 1015$, so $P(M\cap\overline{E})=\frac{1015}{3537}\approx0.287$. ## Step5: Calculate conditional probability (d) Use the formula $P(F|E)=\frac{P(F\cap E)}{P(E)}$. Since $P(F\cap E)=\frac{700}{3537}$ and $P(E)=\frac{795}{3537}$, then $P(F|E)=\frac{700}{795}\approx0.881$. ## Step6: Check for dependence (e) Two events $A$ and $B$ are independent if $P(A\cap B)=P(A)\times P(B)$. $P(F)=\frac{2427}{3537}$, $P(E)=\frac{795}{3537}$, $P(F\cap E)=\frac{700}{3537}$. $P(F)\times P(E)=\frac{2427}{3537}\times\frac{795}{3537}\approx0.153\neq\frac{700}{3537}\approx0.198$. So the events female and education major are dependent. # Answer: a. $\frac{2427}{3537}\approx0.686$ b. $\frac{2522}{3537}\approx0.713$ c. $\frac{1015}{3537}\approx0.287$ d. $\frac{700}{795}\approx0.881$ e. Dependent