Consider randomly selecting a single individual and having thatperson test drive 3 di? erent vehicles. De? ne events A1, A2, and A3, byA1 = likes vehicle #1, A2 = likes vehicle #2, A3 = likes vehicle #3. Suppose that P (A1) = 0.55, P (A2) = 0.65, P (A3) = 0.70, P (A1? A2) = 0.80, P (A2 n A3) = 0.40, and P (A1? A2? A3) = 0.88. a. What is the probability that the individual likes both vehicle #1and vehicle #2? b. Determine and interpret P (A2 |A3). c. Are A2 and A3 independent events? Answer in two di? erent ways. d. If you learn that the individual did not like vehicle #1, what nowis the probability that he/she liked at least one of the other twovehicles?

a) The probability that the individual like both vehicle no. 1 and vehicle no. 2 is 0.40.

b) P (A2|A3) = 0.57

c) A2 and A3 are not independent events

d) Probability that the individual liked the other two vehicles given that he liked the first one is 0.88.

Step-by-step explanation:

P (A1) = 0.55

P (A2) = 0.65

P (A3) = 0.70

P (A1 U A2) = 0.80

P (A2 ∩ A3) = 0.40

P (A1 U A2 U A3) = 0.88

a) We need to find P (A1 ∩ A2). For this we will use the equation:

P (A1 U A2) = P (A1) + P (A2) - P (A1 ∩ A2)

P (A1 ∩ A2) = P (A1) + P (A2) - P (A1 U A2)

= 0.55 + 0.65 - 0.80

P (A1 ∩ A2) = 0.40

The probability that the individual like both vehicle no. 1 and vehicle no. 2 is 0.40.

b) P (A2 |A3) is a conditional probability which can be calculated using the conditional probability formula.

P (A2|A3) = P (A2 ∩ A3) / P (A3)

= 0.40/0.70

P (A2|A3) = 0.57

This is the probability that the individual will like car no. 2 given that he likes car no. 3. P (A2|A3) is read as the probability that A2 will occur given that A3 has already occurred.

c) We need to see if A2 and A3 are independent events. For independent two independent events, the following relation stands true:

P (A and B) = P (A) * P (B)

P (A ∩ B) = P (A) * P (B)

Lets see if this relation stands true for A2 and A3.

We know that P (A2∩A3) = 0.40.

P (A2) * P (A3) = 0.65*0.70

P (A2) * P (A3) = 0.455

We can clearly see that P (A2∩A3) ≠ P (A2) * P (A3) since 0.40 is not equal to 0.455. So it can be concluded that events A2 and A3 are not independent.

Another way to determine this is by considering the fact that if an individual drives three vehicles, the chances that he will like a certain car depends on how much he likes the other two cars. Which means that the probabilities of liking are not independent of each other. In the previous part, we calculated P (A2|A3) which shows that the probability of liking car 2 depends on liking car 3. So, A2 and A3 are not independent events.

d) Now it is given that the individual did not like vehicle no. 1 and we need to find the probability that he liked either only vehicle no. 2 or both vehicle no. 2 and 3. Which means we need to compute P (A2|A1) and P (A2∩A3|A1) and then add both the probabilities to find the final answer.

P (A2|A1) = P (A2 ∩ A1) / P (A1)

we know from part (a) that P (A2 ∩ A1) = 0.40. So,

P (A2|A1) = 0.40/0.55

P (A2|A1) = 0.73

P (A2∩A3|A1) = P (A2∩A3∩A1) / P (A1)

We need P (A2∩A3∩A1), for that we will use the relation:

P (A1UA2UA3) = P (A1) + P (A2) + P (A3) - P (A1∩A2) - P (A1∩A3) - P (A2∩A3) + P (A1∩A2∩A3)

We know all the values except P (A1∩A3). So,

P (A1∩A3) = P (A1|A3) * P (A3)

P (A1|A3) = 1 - P (A2|A3)

= 1 - 0.57

P (A1|A3) = 0.43

P (A1∩A3) = 0.43 * 0.70

P (A1∩A3) = 0.301

P (A1UA2UA3) = P (A1) + P (A2) + P (A3) - P (A1∩A2) - P (A1∩A3) - P (A2∩A3) + P (A1nA2nA3)

0.88 = 0.55 + 0.65 + 0.70 - 0.40 - 0.301 - 0.40 + P (A1∩A2∩A3)

P (A1∩A2∩A3) = 0.081

P (A2∩A3|A1) = P (A2∩A3∩A1) / P (A1)

= 0.081/0.55

P (A2∩A3|A1) = 0.15

Probability that the individual liked the other two vehicles given that he liked the first one is:

P (A2|A1) + P (A2∩A3|A1)

0.73 + 0.15 = 0.88

Probability that the individual liked the other two vehicles given that he liked the first one is 0.88