Question

At a particular garage, 60% of the customers require an oil change, 20% require tire rotation, and 75% of customers require at least one of these services. For a randomly selected customer,

a. What is the probability that the customer will need both services?

b. What is the probability that the customer will need an oil change, but not a tire rotation?

c. What is the probability that the customer will want exactly one of these two services?

Answer 1

a) P(C∩T) = 0.05

b) P(C) = 0.55

c) P(C) + P(T) = 0.70

Explanation:

Given:

60% of the customers require an oil change

20% require tire rotation

75% of customers require at least one of these services.

So,

Let C represent customers that requires oil change and

T represent customers that requires tire rotation.

P(C) + P(C∩T) = 60% = 0.6 .......1

P(T) + P(C∩T) = 20% = 0.2. .......2

P(C) + P(T) + P(C∩T) = 75% = 0.75 .....3

a) the probability that the customer needs both services P(C∩T).

From eqn 1 , 2 , 3 above.

P(C∩T) = eqn 1 + eqn 2 - eqn 3

P(C∩T) = 0.60 + 0.20 - 0.75

P(C∩T) = 0.05 or 5%

b) from eqn 1:

P(C) = 0.60 - P(C∩T)

P(C) = 0.60 - 0.05

P(C) = 0.55 or 55%

c) the probability that a customer will want exactly one of the two services P(C) + P(T).

Adding eqn 1 and 2 we have;

P(C) + P(T) + 2 P(C∩T) = 0.60 +0.20

P(C) + P(T) = 0.80 - 2P(C∩T)

P(C) + P(T) = 0.80 -2(0.05) = 0.70 or 70%

Answer 2

a. 5%

b. 55%

c. 70%

Explanation:

a. The probability of customer wanting both services (P(O&T)) is:

`P(O)+P(T)+P(O\&T) = 0.75\\P(O)+P(O\&T) =0.60\\P(T)+P(O\&T) = 0.20\\P(O)+P(T)+P(O\&T) -[P(O)+P(T)+2P(O\&T)]=0.75 -(0.60-0.20)\\P(O\&T)=0.05=5\%`

The probability is 5%

b. The probability that the customer will need an oil change, but not a tire rotation (P(O)) is :

`P(O)+P(O\&T) = 0.60\\P(O\&T) = 0.05\\P(O) =0.60-0.05 = 0.55 = 55\%`

The probability is 55%

c. The probability that the customer will want exactly one of these two services (P(O)+P(T)) is:

`P(O)+P(T)+P(O\&T) = 0.75\\P(O\&T) = 0.05\\P(O)+P(T) =0.75-0.05 = 0.70 = 70\%`

The probability is 70%