Answer:
If a family chosen at random bought a car, we need to find the probability that the family had not previously indicated an intention to buy a car = P(I'|B) = 0.3362
Step-by-step explanation:
Let the event that a family that intends to buy a car be I
Let the event that a family does not intend to buy a car be I'
Let the event that a family buys a car in those 3 months be B
Let the event that a family does not buy a car in those 3 months be B'
Given,
P(B|I) = 0.70
P(B|I') = 0.10
P(I) = 0.22
P(I') = 1 - P(I) = 1 - 0.22 = 0.78
If a family chosen at random bought a car, we need to find the probability that the family had not previously indicated an intention to buy a car = P(I'|B)
The conditional probability P(A|B), is given as
P(A|B) = P(A n B) ÷ P(B)
So,
P(B|I) = P(B n I) ÷ P(I)
P(B n I) = P(B|I) × P(I) = 0.70 × 0.22 = 0.154
P(B|I') = P(B n I') ÷ P(I')
P(B n I') = P(B|I') × P(I') = 0.10 × 0.78 = 0.078
P(B) = P(B n I) + P(B n I') = 0.154 + 0.078 = 0.232
P(B') = 1 - 0.232 = 0.768
P(I'|B) = P(B n I') ÷ P(B)
= 0.078 ÷ 0.232 = 0.3362
Hope this Helps!!!
