1 A foreman for an injectionmolding firm admits that on 22 o

1) A foreman for an injection-molding firm admits that on 22% of his shifts, he forgets to shut off the injection machine on his line. This causes the machine to overheat, increasing the probability that a defective molding will be produced during the early morning run from 4% to 18%. If a molding is randomly selected from the early morning run of a random day, what is the probability that it is defective?

2)

An aerospace company has submitted bids on two separate federal government defense contracts. The company president believes that there is a 38% probability of winning the first contract. If they win the first contract, the probability of winning the second is 69%. However, if they lose the first contract, the president thinks that the probability of winning the second contract decreases to 55%.

A. What is the probability that they win both contracts?

Probability =

B. What is the probability that they lose both contracts?

Probability =

C. What is the probability that they win only one contract?

Probability =

answer:

answer:

Solution

(1)

Data given is as follows:

P(forget to shut the machine) = 0.22

P(defective mold if machine is always shut properly) = 0.4

P(defective mold | forgot to shut machine) = 0.18

We need to calculate the probability: P(mold is defective)

Using the formula:

P(mold is defective) = P(mold is defective AND machine was forgot to shut) + P(mold is defective AND machine was shut properly)

Now,

P(defective mold | forgot to shut machine) = P(mold is defective AND machine was forgot to shut) / P(forgot to shut)

So,

P(mold is defective AND machine was forgot to shut) = 0.18*0.22 = 0.0396

So,

P(mold is defective) = 0.0396 + 0.04 = 0.0796

(2)

Probability of winning first contract, P(A) = 0.38

Probability of winning second, when first contract is won = P(B|A) = 0.69

Probability of winning second, when first contract is lost = P(B|A\') = 0.55

(a)

Probability of winning both contracts, P(A AND B) = P(B|A)*P(A) = 0.69*0.38 = 0.2622

(b)

P(B) = P(B AND A) + P(B AND A\') = P(B|A)*P(A) + P(B|A\')*P(A\') = 0.2622 + (0.55*(1-0.38)) = 0.6032

So,

Probability of losing both contracts = P(A\' AND B\') = P(A\')*P(B\'|A\') = (1-0.38)*(1-0.55) = 0.279

(c)

Probability of winning only one contract = P(A AND B\') + P(A\' AND B) = P(B\'|A)*P(A) + P(B|A\')*P(A\') = (1-0.69)*0.38 + (0.55*(1-0.38)) = 0.4588