Question

The following table shows the breakdows of opinions for both faculty and students in a recent survey about the new restructuring of the campus to include an HonomCollege. Find the probability that a randomly selected person is either a student opposed to the change or a faculty member who has an opinion either for or againstExpress your answer as a fraction in lowest terms or a decimal rounded to the nearest millionth,

Answer 1

Given,

The number of total students and faculty is 213.

From the table,

The number of students opposed to the change is, n(student oppose) = 44.

The number of faculty members opposed to the change is, n(faculty oppose) = 20.

The number of faculty members favor to the change is, n(faculty favour) = 11.

The probability of elected person is a student oppose to the change or faculty member either for or against is,

`\begin{gathered} \text{Probability}=\frac{n\mleft(\text{student oppose}\mright)+n\mleft(faculty\text{ oppose}\mright)+n\mleft(faculty\text{ favour}\mright)}{\text{total members}} \\ =(44+20+11)/(213) \\ =(75)/(213) \\ =(25)/(71) \\ =0.352113 \end{gathered}`

Hence, the probability is 25/71.