Question

a lot of 1115 components contains 275 that are defective. two components are drawn at random

Problem No. 5.4 110 pt. A lot of 1115 components contains 275 that are defective. Two components are drawn at random and tested. Let A be the event that the first component drawn is defective, and let B be the event that the second component dràwn is defective. 1) Find P(A). 2) Find P(BIA). 3) Find P(AnB) 4) Find P(ACnB) 5) Find P(B). 6) Find P(AB). 7) Are A and B independent? Is it reasonable to treat A and B as though they were independent? Explain. Show your explanations. Displaying only the final answer is not enough to get credit. Note: round calculated numerical values to the fourth decimal place where applicable.

Answer 1

It has been given that

A will be event when first component is drawn & it will be defective

B will be event when 2nd event drawn & found to be defective

Also given

a lot of 1115 components contains 275 that are defective. two components are drawn at random

Probability of event A will be

$P(A)=\frac{275}{1115}=0.2466$

2)

We have to find P(B/A)

Which means that second sample drawn will be defective which is indicated by B & given that first component is defective

So Total Components will be left are 1114 and also defective items will be 274

$P(B/A)=\frac{274}{1114}=0.2459$

3)

We have to find $P(A\cap B)$ Both components drawn will be defective

$P(A\cap B)=P(B/A)\times P(A)=\frac{274}{1114}\times \frac{275}{1115}=0.06066$

4)

We have to find the

$P(A^{c}\cap B)=P(B/A^{c})\times P(A^{c})=P(B/A^{c})(1-P(A))=0.24685(1-0.2466)=0.2468\times 0.7533=0.1859$

Let us find the probability

$P(B/A^{c})$

which means that probability that second part is defective given that first part is not defective

after the first component drawn and it will not be defective

So there will be 1114 Components will be left and defective items will be 275

So $P(B/A^{c})=\frac{275}{1114}=0.24685$

5)

We have to calculate P(B)

By using the law of probability

$P(B)=P(A\cap B)+P(A^{c}\cap B)=0.06066+0.1859=0.2466$

6)

$P(A/B)=\frac{P(B/A)P(A)}{P(B)}=\frac{0.2459\times 0.2466}{0.2466}=0.2459$

7)

Events A and B are not independent

They are dependent of each other

Because $P(B/A)\neq P(B)$