Question

A lot of 20 components contains 4 that are defective. Two components are drawn at random and tested. Let A be the even that the first component drawn is defective and let B be the event that the second component drawn is defective.
(A) find P( A)
(B) find P(B|A)
(C) P( A∩ B)
(d) find P( A^c ∩ B)
(e) find P (A ∩ B^c)
(f) Find P(A^c ∩ B^c)
(g) find P(B)
(H) Find P( A U B)

Answer 1

Okay Finding P(A) is easy.

It is the possibility of getting a defective component from the total number of components. What will that be?

Answer 2

(4/20)

Answer 3

Good work :)

Answer 4

thank you :) I just have trouble with the others.

Answer 5

I haven't learned this topic so this is new to me

Answer 6

Okay for B)

It asks "what is the prob. of B being defective if we know that A was defective?"
The answer is the same as always: favorable / total possibility

=3/19

Answer 7

ok the symbols were confusing. but that makes sense :)

Answer 8

P(B|A) is read as Probability of B given A.

Answer 9

Okay

Answer 10

Shall we move to C?

Answer 11

yes please

Answer 12

C asks for what is the probability that BOTH are defective.

Chance of Event A being defective 3/20
Next event B: 2/19

So simply multiply them to get P(A intersection B)

(3/20)*(2/19)

Answer 13

Wait its 4/20 Sorry for typo

Answer 14

ok but for Even A being defective is 4/20 right?
so ∩ means and?

Answer 15

^^Yes! Happy to see that you understood that.

Answer 16

ok im starting to understand it :)

Answer 17

for D) is it 1-P(A) + P(B)?

Answer 18

Not exactly...Can I leave for dinner? I will get back to you on this.

Answer 19

ok thank you ill be here :)

Answer 20

Great :)..Thanks for understanding.

Answer 21

no problem :D

Answer 22

I am back :)

Answer 23

that was quick :)

Answer 24

Indeed :)

Answer 25

Okay so A^c represent A complement. It is the opposite of what A refers to. So what will be P(A^c) ?

Answer 26

so 1- P(A) ?

Answer 27

Yes That is?

Answer 28

0.8

Answer 29

Good.. Now we take B. What is P(B) Now?

Answer 30

3/19

Answer 31

Exactly! Now here is the important part to understand:

When we have intersection of some events ie both occur together. Then We use AND and for union we use OR. Mathematically AND operation is done by multiplication and OR operation by addition. This is a very important concept.

Answer 32

Now do you understand why I said 1-P(A)+P(B) is wrong?

Answer 33

yup

Answer 34

So can you correct your mistake?

Answer 35

since this is a intersection we need to multiply

Answer 36

Yes so the answer will be (1-P(A))P(B) rather than 1-P(A)+P(B)

Answer 37

:D i understand it.

Answer 38

Good :D

Answer 39

I think with this knowledge you can do the rest of the questions easily! :)

Answer 40

so for the next part it will be P(A) * (1-P(B))

Answer 41

Indeed ;)

Answer 42

Great

Answer 43

Can you do the rest by yourself?

Answer 44

/answer is 16/95

Answer 45

yes i think i can

Answer 46

I dont want numerical answers. I want you to understand the concept. Then you can get all the answers easily.

Answer 47

Good :)

Answer 48

Yes after you explained the concept its easier to do it. I had no idea what the ∩ and U sign meant.

Answer 49

Havent you done the chapter called "Sets"?

Answer 50

I havent got the textbook for this class. So everything is getting harder to understand.

Answer 51

What grade is this for?

Answer 52

10th

Answer 53

Thank you for all your help!! I had no clue how to do this. now i understand :)

Answer 54

If you have time go through this Chapter on Sets. Then move to probability. Both are closely related.

http://www.scs.ryerson.ca/~mth110/Handouts/PD/setbasics.pdf

Answer 55

Thank you I will look through this

Answer 56

You are welcome!