Question

Of all the soft drink consumers in a particular sales region, 30% prefer Brand A and 70% prefer Brand

b. Of all these soft drink consumers, 20% prefer Brand A and are female, and 40% prefer Brand B and are female. What is the probability that a randomly selected consumer is female, given that the person prefers Brand A? 0.18 0.21 0.34 0.67

Answer 1

67% (0.67) because 30% like brand a and 20% of people are females that like brand A so 20:30 is 2/3 which is 0.6 repeating which rounds to 0.67

Answer 2

Option D -0.67

Explanation:

Given : Preference of Brand A =30% ⇒
`P(A)=(30)/(100)=0.3`

Preference of Brand B =70% ⇒
`P(B)=(70)/(100)=0.7`

Prefer Brand A and are Female = 20% ⇒
`P(A/F)=(20)/(100)=0.2`

Prefer Brand B and are Female = 40% ⇒
`P(B/F)=(40)/(100)=0.4`

To find : Selected consumer is female, given that the person prefers Brand A
`P(F/A)`

Solution : Using Bayes' theorem, which state that

`P(A/B)=(P(B/A)P(A))/(P(B))`

where, P(A) and P(B) are probabilities of observing A and B.

P(B/A)= is a conditional probability where event B occur and A is true

P(A/B)= also a conditional probability where event A occur and B is true.

Now, applying Bayes' theorem,

`P(F/A)=(P(A/F)P(A))/(P(B)P(B/F)+P(A)P(A/F))`

`P(F/A)=((0.2)(0.3))/((0.7)(0.4)+(0.3)(0.2))`

`P(F/A)=(0.6)/(0.28+0.6)`

`P(F/A)=(0.6)/(0.88)`

`P(F/A)=0.68`

Therefore, Option D is correct probability that a randomly selected consumer is female, given that the person prefers Brand A -0.67