Question

The members of a consulting firm rent cars from three rental agencies: 50% from agency X, 30% from agency Y and 20% from agency Z. From past experience it is known that 9% of the cars from agency X need a service and tuning before renting, 12% of the cars from agency Y need a service and tuning before renting and 10% of the cars from agency Z need a service and tuning before renting. If the rental car delivered to the firm needs service and tuning, find the probability that agency Z is not to be blamed.

Answer 1

Solution:

Probability of cars from agency X need a service and tuning before renting P(X)=0.09

Probability of cars from agency X not need a service and tuning before renting P(X')=0.91

Probability of cars from agency Y need a service and tuning before renting P(Y)=0.12

Probability of cars from agency Y not need a service and tuning before renting P(Y')=0.79

Probability of cars from agency Z need a service and tuning before renting P(Z)=0.10

Probability of cars from agency Z not need a service and tuning before renting P(Z')=0.90

To find the probability that agency Z is not to be blamed,we had to apply Bay's theorem of Probability

first it should be noted that company Z is not to be blamed,ie we must used P(Z')=0.90,along with that car taken is from agency Z

Probability:
$= \frac{p(z) \times0.20 }{p(x) \times0.50 + p(y) \times0.30 + p(z) \times0.20 } \\ \\ = \frac{0.9 \times 0.2}{0.91 \times 0.5 + 0.79 \times 0.3 + 0.9 \times 0.2} \\ \\ = \frac{0.18}{0.455 +0 .237 + 0.18} \\ \\ = \frac{0.18}{0.872} \\ \\ =0.206$

probability that agency Z is not to be blamed=0.206

Answer 2

Answer: this is the answer. Hope you understand

Step-by-step explanation: