PROBLEM 8E

A store sells four brands of tablets. The least expensive brand, B1, accounts for 40% of the sales. The other brands (in order of their price) have the following percentages of sales: B2, 30%; B3, 20%; and B4, 10%. The respective probabilities of needing repair during warranty are 0.10 for B1, 0.05 for B2, 0.03 for B3, and 0.02 for B4. A randomly selected purchaser has a tablet that needs repair under warranty. What are the four conditional probabilities of being brand Bi, i = 1, 2, 3, 4?

Solution 8E

Step1 of 3:

We have,

Probability that the tablets need repair = P(R)

Probability that the least expensive brands are

P() = 40%

= 0.4

P() = 30%

= 0.3

P() = 20%

= 0.2

P() = 10%

= 0.1

Similarly,

P(R/) = 0.10

P(R/) = 0.05

P(R/) = 0.03

P(R/) = 0.02

We need to find,

What are the four conditional probabilities of being brand Bi, i = 1, 2, 3, 4?

Step2 of 3:

We need to find the Probability that the tablets need repair and it is given by

P(R) = P(R) + P(R) + P(R) + P(R) ………(1)

Where,

1).P(R) = P()P(R/) [Because P(R/) = ]

= 0.40.10

= 0.04

Hence, P(R) = 0.04.

2).P(R) = P()P(R/) [Because P(R/) = ]

= 0.30.05

= 0.015

Hence,P(R) = 0.015.

3).P(R) = P()P(R/) [Because P(R/) = ]

= 0.20.03

= 0.006

Hence, P(R) = 0.006.

4).P(R) = P()P(R/) [Because P(R/) = ]

...