Question

Problem 1:

A Six Sigma Green Belt randomly selects two parts from a box containing 5 defective and 15 nondefective parts. He discards the box if one or both parts are defective. Using what you’re learned in chapter 3, what is the probability that he will select:

1. One defective part

2. Two defective parts

3. Reject the box

Answer 1

GIVEN:

A Six Sigma Green Belt randomly selects two parts from a box containing 5 defective and 15 non defective parts.

SOLUTION:

(a) PROBABILITY THAT HE WILL SELECT ONE DEFEECTIVE PART:

Total number of ways of selecting two parts from a box containing 20 parts (5 defective + 15 non defective) is .

Let A be the event that he will select one defective part.

Number of ways of selecting one defective part from a box containing 20 parts (5 defective + 15 non defective) is (Since he selects one from 5 defective parts and one from 15 non defective parts.)

The formula for probability is given by,

Probability = Number of favourable cases / Total number of cases

Thus,

Probability that he will select one defective part = Number of ways of selecting one defective part from a box containing 20 parts / Total number of ways of selecting two parts from a box

(Since )

  

The probability that he will select one defective part is .

(b) PROBABILITY THAT HE WILL SELECT TWO DEFECTIVE PARTS:

Total number of ways of selecting two parts from a box containing 20 parts (5 defective + 15 non defective) is .

Let B be the event that he will select two defective parts.

Number of ways of selecting two defective parts from a box containing 20 parts (5 defective + 15 non defective) is (Since he selects two from 5 defective parts and none from 15 non defective parts.)

The formula for probability is given by,

Probability = Number of favourable cases / Total number of cases

Thus,

Probability that he will select two defective parts = Number of ways of selecting two defective parts from a box containing 20 parts / Total number of ways of selecting two parts from a box

  

The probability that he will select two defective parts is .

(c) PROBABILITY THAT HE WILL REJECT THE BOX:

He discards the box if one or both parts are defective.

That is,

Let A be the event that he will select one defective part and the probability that he will select one defective part is

Let B be the event that he will select two defective parts and the probability that he will select two defective parts is

  

Let be the event that both the parts are defective and the probability of both parts being defective by applying the formula as follows:

Total number of parts = 20

Number of defectives = 5

Consider A to be the event that the first selected part is defective

Therefore, the probability of selecting the 1^st defective part can be shown as follows:

  

Now, let us compute the conditional probability of the second selected part being defective provided that first selected part is defective as follows:

Thus the probability of both parts being defective is,

Thus the probability that he reject the box is given by,

Thus the probability of rejecting the box is .