Question

Consider randomly selecting a single individual and having that person test

drive 3 different vehicles. Define events A1, A2, and A3 by A1=likes vehicle #1,

A2 =likes vehicle #2, A3 =likes vehicle #3. Suppose that ( ) 65.0 P A1 = ,

( ) 55.0 P A2 = , ( ) 70.0 P A3 = , ( ) 80.0 P A1 ∪ A2 = , ( ) ,60.0 P A1 ∪ A3 = ( ) 40.0 P A2 ∪ A3 = ,

( ) 88.0 P A1 ∪ A2 ∪ A3 = .

a. What is the probability that the individual likes both vehicle #1 and vehicle #3?

b. Determine and interpret ( / ) P A1 A3.

c. Are A1 and A2 independent events? Answer in two different ways.

d. If you learn that the individual did not like vehicle #2, what now is the

probability that he/she liked at least one of the other two vehicles

Answer 1

QUESTION

Consider randomly selecting a single individual and having that person test

drive 3 different vehicles. Define events A1, A2, and A3 by A1 = 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 ∩ A3) = 0.50, and P(A1 ∪ A2 ∪ A3) = 0.88.

(a) What is the probability that the individual likes both vehicle #1

and vehicle #2?

(b) Determine and interpret P (A2 |A3 ).

(c) Are A2 and A3 independent events? Answer in two different ways.

(d) If you learn that the individual did not like vehicle #1, what now

is the probability that he/she liked at least one of the other two

vehicles?

SOLUTION

The events are defined as follows:

A₁ = an individual like vehicle #1

A₂ = an individual like vehicle #2

A₃ = an individual like vehicle #3

The information provided is:

$P(A_1)=0.55\\P(A_2)=0.65\\P(A_3)=0.70\\P(A_1\bigcup A_2)=0.80\\P(A_2\bigcap A_3)=0.50\\P(A_1\bigcup A_2A_3)=0.88$

Question

(a) Compute the probability that the individual likes both vehicle #1 and vehicle #2 as follows:

Solution

$P(A_1\bigcap A_2)=P(A_1)+P(A_2)-P(A_1\bigcup A_2)\\=0.55+0.65-0.80\\Answer=0.40$

Question

(b) Determine and interpret P (A2 |A3 ).

Solution

$P(A_2|A_3)=\frac{P(A_2\bigcap A_3)}{P(A-3)}\\=\frac{0.50}{0.70}\\Answer=0.7143$

Question

(c) Are A2 and A3 independent events? Answer in two different ways.

Solution

If two events X and Y are independent then;

$P(X\bigcap Y)=P(X)*P(Y)\\P(X|Y)=P(X)$

The value of P (A₂ ∩ A₃) is 0.50.

The product of the probabilities, P (A₂) and P (A₃) is:

$P(A_2)*P(A_3)=0.65*0.70=0.455$

Thus, P (A₂ ∩ A₃) ≠ P (A₂) × P (A₃)

The value of P (A₂ | A₃) is 0.7143.

The value of P (A₂) is 0.65.

Thus, P (A₂ | A₃) ≠ P (A₂).

The events A₂ and A₃ are not independent.

Question

(d) If you learn that the individual did not like vehicle #1, what now is the probability that he/she liked at least one of the other two vehicles?

Solution

$P(A_2 \bigcup A_3|A^C_1)=\frac{P((A_2\bigcup A_3)\bigcap A^C_1)}{P( A^C_1)}\\=\frac{0.88-0.55}{1-0.55}\\=0.7333$

Thus, the probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333.