Question

a local high school has both male and female students. each student either plays a sport or doesnt. the two - way table summarizes a random sample of 80 students.

	female	male
sport	36	17

let sport be the event that a randomly chosen student (from the table) plays a sport. let female be the event that a randomly chosen student (from the table) is female. find the following probabilities. write your answers as decimals.
(a) ( p(\text{female}) = square )
(b) ( p(\text{sport and female}) = square )
(c) ( p(\text{sport }| \text{ female}) = square )

Explanation:

Step1: Calculate total number of students

First, we find the total number of students by adding up all the values in the two - way table. The number of female students who don't play sport is 12, female students who play sport is 36, male students who don't play sport is 15, and male students who play sport is 17. So the total number of students \(n = 12 + 36+15 + 17=80\).

Step2: Calculate \(P(\text{female})\)

The number of female students is the sum of female students who play sport and who don't play sport. So the number of female students \(n(\text{female})=12 + 36 = 48\). The probability \(P(\text{female})\) is the number of female students divided by the total number of students. So \(P(\text{female})=\frac{n(\text{female})}{n}=\frac{48}{80}=0.6\).

Step3: Calculate \(P(\text{sport and female})\)

The number of students who are female and play sport is 36. So \(P(\text{sport and female})=\frac{\text{number of students who are female and play sport}}{\text{total number of students}}=\frac{36}{80} = 0.45\).

Step4: Calculate \(P(\text{sport }|\text{ female})\)

By the formula for conditional probability, \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Here \(A=\text{sport}\) and \(B = \text{female}\). We know that \(P(A\cap B)=P(\text{sport and female}) = 0.45\) and \(P(B)=P(\text{female})=0.6\). So \(P(\text{sport }|\text{ female})=\frac{0.45}{0.6}=0.75\). We can also calculate it as the number of female students who play sport divided by the number of female students. The number of female students who play sport is 36, and the number of female students is 48. So \(\frac{36}{48}=0.75\).

Answer:

(a) \(P(\text{female}) = 0.6\)
(b) \(P(\text{sport and female})=0.45\)
(c) \(P(\text{sport }|\text{ female}) = 0.75\)