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

BOf 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

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jakedc0614 jakedc0614 · Answer 1 · 2015-07-13T22:58:09-04:00

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

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tardymanchester tardymanchester · Answer 2 · 2019-01-21T03:50:50-05:00

Answer:

Option D -0.67

Step-by-step explanation:

Given : Preference of Brand A =30% ⇒ \(P(A)=\frac{30}{100}=0.3\)

Preference of Brand B =70% ⇒ \(P(B)=\frac{70}{100}=0.7\)

Prefer Brand A and are Female = 20% ⇒ \(P(A/F)=\frac{20}{100}=0.2\)

Prefer Brand B and are Female = 40% ⇒ \(P(B/F)=\frac{40}{100}=0.4\)

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

Solution : Using Bayes' theorem, which state that

\(P(A/B)=\frac{P(B/A)P(A)}{P(B)}\)

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,

\(P(F/A)=\frac{P(A/F)P(A)}{P(B)P(B/F)+P(A)P(A/F)}\)

\(P(F/A)=\frac{(0.2)(0.3)}{(0.7)(0.4)+(0.3)(0.2)}\)

\(P(F/A)=\frac{0.6}{0.28+0.6}\)

\(P(F/A)=\frac{0.6}{0.88}\)

\(P(F/A)=0.68\)

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