...

Question

...

anonymous · Answer

for the first to be defective, it will be $$1-(\frac{ 9 }{ 12 })^{2}$$?

anonymous · Answer

i meant 9/12 raised to the 9th power

ganeshie8 · Answer

nope it should be simply  
$$\dfrac{3}{12}$$

anonymous · Answer

for part b, it would be 3/12 * 9/11?

ganeshie8 · Answer

Yep

anonymous · Answer

for both to be defective, it would be 3/12 * 2/11?

ganeshie8 · Answer

You got it!

anonymous · Answer

for the second to be defective, it would be.......

anonymous · Answer

2/11?

ganeshie8 · Answer

d and e are tricky

anonymous · Answer

i figured lol

ganeshie8 · Answer

`d)	The second air conditioner is defective `

forget about numbers and lets see in what ways this event can ever happen

anonymous · Answer

in part d, we have to assume the probability of the first to be of either defective or good?

ganeshie8 · Answer

Exactly!

ganeshie8 · Answer

just add up the probabilities : 

P( `first defective`   and  `second defective`)

+

P( `first good`   and  `second defective`)

anonymous · Answer

oh, i was thinking of subtracting something as well from that

ganeshie8 · Answer

thats good to think but i don't see anything to subtract here hmm

anonymous · Answer

because they are not independent of each other, correct?

anonymous · Answer

i meant dependent...

ganeshie8 · Answer

i guess so..

anonymous · Answer

and for exactly one to be defective, it would be
$$1 - (\frac{ 9 }{ 12 })^{9}$$

anonymous · Answer

We disregard the probability of choosing a good event.

ganeshie8 · Answer

thats a good try but i think it is wrong

ganeshie8 · Answer

keep in mind you're looking at the parts one by one

ganeshie8 · Answer

i.e, with out replacement

anonymous · Answer

would it be the summation of the first to be defective and the second to be defective?

ganeshie8 · Answer

`e)	Exactly one is defective `

we can have this in two ways : 

P( `first defective` and  `second good`)

+

P( `first good` and `second defective`)

ganeshie8 · Answer

you get 
$$\dfrac{3}{12}\times \dfrac{9}{11}  + \dfrac{9}{12}\times \dfrac{3}{11}$$

anonymous · Answer

But that means we have possibilities of good events?

anonymous · Answer

for what i said, it would be equal to:
P(first is defective) + P(second is defective) - P(both are defective)
But i like yours better.

ganeshie8 · Answer

I see what you're doing now :) that looks good