Question

A School of Public Health completed a study on alcohol consumption on college campuses. They concluded that 20.620.6​% of women attending​ all-women colleges abstained from​ alcohol, compared to 17.317.3​% of women attending coeducational colleges. Approximately 4.64.6​% of women college students attend​ all-women schools. Complete parts​ (a) and​ (b) below.

(a) What is the probability that a randomly selected female student abstains from​ alcohol?

​(Type an integer or a decimal rounded to four decimal places as​ needed.)

(b) If a randomly selected female student abstains from alcohol, what is the probability she attends a coeducational college?

​(Type an integer or a decimal rounded to four decimal places as​ needed.)

Answer 1

Solution:-

let P(x) represent women attending all women college , P(x)=0.046

P(y) represent women attending coeducational, P(y)=1-0.046=0.954

P(z) represent probability of abstained from alcohol

Given,

Women attending all women colleges, probability of abstained from alcohol :P(z/x)=0.206

Women attending coed colleges, probability of abstained from alcohol: P(z/y)=0.173

a.)

Probability that a randomly selected female student abstains from​ alcohol=P(z)

P(z)=P(z/x)*P(x)+P(z/y)*P(y)

=0.206*0.046+0.173*0.954

=0.1745 ......ans

b.)

Randomly selected female student abstains from alcohol, probability she attends a coeducational college= P(y/z)

P(y/z)=P(z/y)*P(y)/P(z)

=0.173*(0.954/0.1745)

=0.9458 ....ans