Question

if 20 percent of the bolts produced in a machine are defective, find the probability that out of 4 bolts chosen at random (1) one(2)none(3)less than two will be defective​

Answer 1

Answer:

Step-by-step explanation:

Given, n=10,p=  

100

20

​  

=  

5

1

​  

 

∴q=1−  

5

1

​  

=  

5

4

​  

 

Let X denote the number of defective bolts chosen.

∴X=2

(i) Using Binomial distribution

P(X−2)=10C  

2

​  

 

(  

5

1

​  

)  

2

⋅(  

5

4

​  

)  

8

=  

1×2

10×9

​  

(  

5  

10

 

4  

8

 

​  

)=45(  

5  

10

 

4  

8

 

​  

)

(ii) Using Poisson distribution

λ=np=10⋅(  

5

1

​  

)=2

P(X−x)=e  

−λ

 

∠x

λ  

x

 

​  

,x=0,1,2,....

P(X=2)=e  

−2  

∠2

2  

2

 

​  

=e  

−2

[  

1×2

4

​  

]=e  

−2

(2)

=2(0.1353)=0.2706.