Question

At a college, $72\%$of courses have final exams and $46\%$of courses require research papers. Suppose that $32\%$of courses have a research paper and a final exam. Let F be the event that a course has a final exam. Let R be the event that a course requires a research paper.

a. Find the probability that a course has a final exam or a research project.

b. Find the probability that a course has NEITHER of these two requirements.

Short answer

(a) The probability that a course has a final exam or a research project is $0.86$.

(b) The probability that a course has neither a final exam nor a research project is$0.14$.

Step-by-step solution

Given information (part a)

At a college, $72\%$of courses have final exams and $46\%$of courses require research papers. Suppose that $32\%$of courses have a research paper and a final exam.

Explanation (part a)

Let the events be

$F=$Event that a course has a final exam

$R=$Event that a course has a research paper

We have

localid="1648100323804" $$\begin{gathered} P\left(F\right)=0.72 \\ P\left(R\right)=0.46 \\ P\left(FandR\right)=0.32 \end{gathered}$$

We need to calculate the probability that a course has a final exam or a research project

Thus is given as $P\left(FORR\right)$

localid="1648100352897" $P(ForR)=P\left(F\right)+P\left(R\right)-P(FandR)$

Substituting the values, we get

localid="1648100380895" $$\begin{gathered} P\left(F\text{or}R\right)=0.72+0.46-0.32 \\ P(ForR)=1.18-0.32 \\ P(ForR)=0.86 \end{gathered}$$

Given information (part b)

At a college, $72\%$of courses have final exams and $46\%$of courses require research papers. Suppose that $32\%$of courses have a research paper and a final exam.

Explanation (part b)

Let the events be

$F=$Event that a course has a final exam

$R=$Event that a course has a research paper

We have

$$\begin{gathered} P\left(F\right)=0.72 \\ P\left(R\right)=0.46 \\ P\left(FandR\right)=0.32 \\ P\left(ForR\right)=0.86 \end{gathered}$$

We need to calculate the probability that a course has neither a final exam nor a research project.

Thus is given as $P\left(ForR\right)\text{'}$

$P\left(ForR\right)\text{'}=1-P\left(ForR\right)$

Substituting the values, we get

$$\begin{gathered} P\left(ForR\right)\text{'}=1-0.86 \\ P\left(ForR\right)\text{'}=0.14 \end{gathered}$$