Question

. Assume that a factory has two machines. Past records show that machine 1 produces 30% of all the items of output and machine 2 produces 70% of the items. Further 5% of the items produced by machine 1 were defective and 1% produced by machine 2 were defective. If a defective item is drawn randomly what is the probability that it is produced by a. machine 1, b. machine 2​

Answer 1

Answer:

Step-by-step explanation:

Let E  

1

​  

 and E  

2

​  

 be the respective events of items produced by machines A and B. Let X be the event that the produced item was found to be defective.

∴ Probability of items produced by machine A,  P(E  

1

​  

)=60% =  

5

3

​  

 

Probability of items produced by machine B,  P(E  

2

​  

)=40%  =  

5

2

​  

 

Probability that machine A produced defective items, P(X∣E  

1

​  

)=2%  =  

100

2

​  

 

Probability that machine B produced defective items, P(X∣E  

2

​  

)=1%  =  

100

1

​  

 

The probability that the randomly selected item was from machine B, given that it is defective, is given by P(E  

2

​  

∣X).

By using Baye's theorem, we obtain

P(E  

2

​  

∣X)=  

P(E  

1

​  

)⋅P(X∣E  

1

​  

)+P(E  

2

​  

)⋅P(X∣E  

2

​  

)

P(E  

2

​  

)⋅P(X∣E  

2

​  

)

​  

 

=  

5

3

​  

⋅  

100

2

​  

+  

5

2

​  

⋅  

100

1

​ 5 /2 -100 /1

​=  5006 +  500 /2  

500 /2

​  =  8 /2

​  

=  4 /1

​  

=0.25

Answer 2

Answer:

Probability of getting a defective item that is produced by M1 = 15/22

Probability of getting a defective item that is produced by M2 = 7/22

Explanation:

Let total output be 100 units

M1 produces 30% of units =

$\frac{30}{100} \times 100 = 30$

m1 produces 30 units

Similarly, M2 produces 70% of the units

$= \frac{70}{100} \times 100 = 70$

M2 produces 70 units

Now,

5% of units produced by M1 are defective

so 5% of 30 units are defective

$= \frac{5}{100} \times 30 = \frac{3}{2}$

similarly,

1% of units produced by M2 are defective

so 1% of 70 units are defective

$\frac{1}{100} \times 70 = \frac{7}{10}$

Probability= possible outcomes/ total number of outcomes

total outcomes =

$\frac{3}{2} + \frac{7}{10} = \frac{15 + 7}{10} = \frac{22}{10}$

Probability of defective items that is produced by M1 =

$\frac{ \frac{3}{2} }{ \frac{22}{10} } = \frac{3}{2} \times \frac{10}{22} = \frac{30}{44} = \frac{15}{22}$

Probability of defective items that is produced by M2 =

$\frac{ \frac{7}{10} }{ \frac{22}{10} } = \frac{7}{10} \times \frac{10}{22} = \frac{70}{220} = \frac{7}{22}$