Question

5) A lumber grading system is known to be defective 20% of the time. When the machine works properly, 20% of the lumber that passes through is rejected, but 50% of the pieces are rejected when it is defective,. In either case, pieces that are accepted are graded this way:

20% #1 Grade

70% #2 Grade

10% #3 Grade

a. Draw a tree diagram outlining each possible outcome. Include all of the

relevant probabilities as well as the probability for each outcome.

b. What is the overall probability of accepting a piece of lumber?

c. What is the overall probability of getting a #2 Grade piece?

d. What is the probability of accepting a piece of lumber if you know that the machine is working properly?

e. What is the probability that a piece that was rejected came from a defective machine?

f. What is the probability that a piece that was rejected came from a working machine?

Answer 1

Solution

Tree diagram represents the intersection and the conditional probability. The first node is the probability of one event happening. The 2nd node is the probability of another event happening given that the first event has already happened. It gives the conditional probability.

By multiplying the probabilities of the 2 nodes we can get the intersection probabilities.

So

Let Variable 'D' denote machine is defective and 'ND' working. Let 'R' denote machine is rejected and 'NR' denote accepted.

E. In the diagram in the first we decide whether the machine is defective or not.

P(D) = 0.2 therefore P(ND) = 0.8 (1-0.2)

In the second node we decided after seeing that whether the machine is defective or not it should be rejected or not. Therefore rejection or acceptance is conditional on defective or working.

Probability of being rejected after passing through means P(R| ND) = 0.2 and

Probability of being rejected after being defective P(R| D) = 0.5.

(a)

Here is the diagram with all the individual , conditional and intersection probabilties.

(b)

Probability of accepting a lumber mean probability that lumber is not rejected. P(NR)

Not rejected follows 2 paths. Either the machine is not defective and not rejected or it is defective and not rejected.

= 0.2 * 0.5 + 0.8 * 0.8 .......................From the diagram

= 0.74

Probability of accepting a lumber is

(c)

Probability of getting #2 Grade lumber.

First we'll have to accept the the lumber and then join the probability of it being #2 Grade.

Ans = P( NR) * P( #2 grade | NR) ...................P(grade 2 after accepting) = 0.7 from the question.

= 0.74 * 0.7

= 0.518

Probability of getting #2 Grade lumber is .

d)

Probability of accepting a lumber given machine is working.

P( NR | ND) = 0.8

e)

Probability that a lumber is rejected given it defective

P( R | D) = 0.5

f)

Probability that a lumber is rejected given it working (Not defective)

P( R | ND) = 0.2