Question

A company manufactures tires at a cost of $40 each. The following are probabilities of defective tires in a given production run: probability of no defective tires: 10%; probability of 1% defective tires: 30%; probability of 2% defective tires: 40%; probability of 3% defective tires: 20%. When a defective tire is detected, it must be removed from the assembly line and replaced. This process adds an extra $5 to the cost of the replaced tire. What is the expected cost to the company of a batch of 3000 tires?

Answer 1

Answer:

The expected cost of the company for a 3000 tires batch is $120255

Step-by-step explanation:

Recall that given a probability of defective tires p, we can model the number of defective tires as a binomial random variable. For 3000 tires, if we have a probability p of having a defective tire, the expected number of defective tires  is 3000p.

Let X be the number of defective tires. We can use the total expectation theorem, as follows: if there are $A_1,\dots, A_n$ events that partition the whole sample space, and we have a random variable X over the sample space, then

$E(X) = p(A_1)E(X|A_1) + \dots + p(A_n)E(X|A_n)$.

So, in this case, we have the following

$E(X) = P(p=0\%)E(X|p=0\%)+P(p=1\%)E(X|p=1\%)+P(p=2\%)E(X|p=2\%)+P(p=3\%)E(X|p=3\%) = 0.1\cdot 3000\cdot 0 + 0.3\cdot 3000\cdot 0.01+ 0.4\cdot 3000\cdot 0.02+ 0.2\cdot 3000\cdot 0.03 = 51$.

Let Y be the number non defective tires. then X+Y = 3000. So Y = 3000-X. Then E(Y) = 3000-E(X). Then, E(Y) = 2949.

Finally, note that the cost of the batch would be 40Y+45X. Then

$E(40Y+45X) = 40E(Y)+45E(X) = 40\cdot 2949+45\cdot 51=120255$