Question

Components of a certain type are shipped to a supplier in batches of ten. Suppose that 49% of all such batches contain no defective components, 27% contain one defective component, and 24% contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with 0, 1, and 2 defective components being in the batch under each of the following conditions? (Round your answers to four decimal places.)

(a) Neither tested component is defective.

no defective components
one defective component
two defective components

(b) One of the two tested components is defective. [Hint: Draw a tree diagram with three first-generation branches for the three different types of batches.]

no defective components
one defective component
two defective components

Answer 1

Answer:

a)

Given,

To determine the required probabilities

P(x) = 49% = 0.49

P(y) = 27% = 0.27

P(z) = 24% = 0.24

let us assume A be the event that neither of the component is defective

This occasion can occur in the accompanying ways:

(i)

There are 0 defects,

at that point probability of getting a non imperfection = 10/10 = 1

The general probability = 0.49 * 1

= 0.49

(ii)

The group has 1 defect , yet our 2 examples misses the imperfection.

Great results, picking 2 out of the 9 non defects = 9C2

= 9*8 / 2

= 36

Absolute Outcomes = picking any 2 out of 10 = 10C2

= 10*9 / 2

= 45

Along these lines the probability = 36/45

= 0.8

Along these lines probability of this occasion happening = 0.27 * 0.8

= 0.216

(iii)

The clump has 2 defects, however our 2 examples misses the defect.

Positive results, picking 2 out of the 8 non faulty = 8C2

= 8*7 / 2

= 28

All out Outcomes = picking any 2 out of 10 = 10C2

= 10*9/2

= 45

Along these lines the probability = 28/45

= 0.622

Thus the probability of this occasion happening = 0.24 * 0.6222

= 0.1493

Along these lines P(A) = 0.49 + 0.622 + 0.1493

P(A) = 0.8553

(a)

Neither tested component is defective for

0 faulty parts ( 0 defects) = P(x/A)

= P(x A)/P(A)

= 0.49 / 0.8553

Probability of no defective components = 0.5729

1 faulty part (1 defective component) = P(y/A)

= P(y A)/P(A)

= 0.622/0.8553

1 defective component = 0.7272

2 faulty part (2 defective) = P(z/A)

= P(z A)/P(A)

= 0.1493 / 0.8553

For 2 defective components = 0.1746

b)

Let the assume that A1 be the one tested component is defective

This occasion can occur in the accompanying ways:

(i)

There are 0 defects, at that point probability of getting an imperfection is = 0

The general probability = 0.49 * 0 = 0

(ii)

The bunch has 1 imperfection , and we get 1 deformity and 1 non deformity.

Ideal occasion: Choosing 1 out of 9 non imperfections and 1 out of the 1 deformity

= 9C1 x 1C1

= 9 x 1

= 9

All out Outcomes = picking any 2 out of 10

= 10C2

= 10*9/2

= 45

Along these lines the probability = 9/45

= 1/5

= 0.2

Along these lines probability of this occasion happening = 0.27 * 0.2

= 0.054

(iii)

The group has 2 imperfections, we get 1 deformity out of the 2.

Positive occasion:

Choosing 1 out of 8 non defects and 1 out of the 2 deformity = 8C1 x 2C1

= 8 x 2

= 16

All out outcomes = picking any 2 out of 10

= 10C2

= 10*9/2

= 45

Along these lines the likelihood = 16/45

= 0.3556

Thus the probability of this occasion happening = 0.24 * 0.3556

= 0.0853

Along these lines P(A1) = 0 + 0.054+ 0.0853

P(A1) = 0.1393

(b)

Given that one tried segment is damaged and the other is non blemished for

0 faulty parts = P(x/A1)

= P(x A1)/P(A1)

= 0 / 0.1393

0 defective component = 0.0000

1 faulty part = P(y/A1)

= P(y A1)/P(A1)

= 0.054 / 0.1393

1 defective component = 0.3877

2 faulty part = P(z/A1)

= P(z A1)/P(A1)

= 0.0853 / 0.1393

2 defective component = 0.6123