Hi rosalie;

I know this particular problem and you have not copied it exactly. Please check the question again. Copying the problem correctly is important.

You should be able to get 1) right away. Please look at the problem again.

2)

This is a conditional probability:
Lets start with this:

P(D) = .83

P(A) = .82

P( D and A ) = .78

Do you remember the formula for a conditional probability?

Hi rzaidan;

That's correct. I made an arithmetic mistake, .95 is right.

Hi rosalie;

Have you looked at my post. Both posts have answered your problem.

Hi rosalie;

Have you looked at my post? That is all you need . You just plug into the formula that I have put in the hidden box.

hi rosalie

Have a look at my picture below.

Imagine 10000 flights.

Split into 8300 that depart on time and 1700 that depart late.

Then put in the 7800 that depart on time and arrive on time.

You can work out that 500 out of 8300 must have departed on time but arrived late.

Altogether 8200 arrive on time.  Take away the 7800 and you are left with

400 that depart late and arrive on time.

Finally there must be 1300 left that depart late and arrive late.

To answer the question "departs on time given it arrives on time" :

There are 8200 flights that arrive on time.

Of these 7800 depart on time.

So P(Depart on time given that it has arrived on time) = 7800/8200.

Hope that helps,

Bob

Uploaded Images

Last edited by bob bundy (2011-03-04 08:16:28)