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**rosalie****Member**- Registered: 2010-05-13
- Posts: 3

Hi, all

I m not being able to solve the following problem.Please help.

The probability that a regular scheduled flight depart on time is 0.83 and the probability that it arrives on time is 0.82 and the probability that it departs and arrives on time is 0.78.Find the probability that a plane

(1) Arrives on time

(2) Depart on time given that it has arrived on time

Please elaborate.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,245

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?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**rzaidan****Member**- Registered: 2009-08-13
- Posts: 59

Dear rosalie

1) P(Arrive only)= P(A-D)=P(A)_P(A and D)

=.82-.78=.04

2)P(D/A)=P(D and A)/P(A)=.78/.82=0.95

I hope I am right

Best Wishes

Riad Zaidan

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,245

Hi rzaidan;

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

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**rosalie****Member**- Registered: 2010-05-13
- Posts: 3

Hi, rzaidan

Thanks, for giving me the solution.But, I didn't get ur point.Could u please elaborate it.I m new to probability so I need explanation.

Please help.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,245

Hi rosalie;

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

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**rosalie****Member**- Registered: 2010-05-13
- Posts: 3

Hi, all

Please solve my problem.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,245

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.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,245

I think she wants to be explained what probability is.

I think that would take many books and probably a couple of teachers or a real urgent need like a hobby problem.

Remember a kid in my class. You know could not do simple arithmetic. Everyone thought he was the dumbest kid in the class. Years later I run in to him and he takes me to the track. It seems all on his own he learned not only arithmetic but how to understand and compute probabilities. He was better than I am. You see,

Necessity is the mother of all inventions.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,191

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

*Last edited by bob bundy (2011-03-03 09:16:28)*

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