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**agus****Member**- Registered: 2011-10-04
- Posts: 2

Hi All,

I have a homework on probability. It goes like this

The probability that 1 percent of the blowout preventers produced by company PreBlow are defective is 0.80, the probability that 5 percent of the preventers are defective is 0.10, and the probability that 10 percent of the preventers are defective is 0.10. A blowout preventer is randomly chosen, tested and is defective.

a. What is the probability that 1 percent of the preventers are defective? That 5 percent are defective? That 10 percent are defective?

b. Suppose that a second preventer is randomly chosen from the companys output, and it too is defective. Following this second observation, what are the probabilities that 1, 5, and 10 percent, respectively, of the items produced by the process are defective?

My answer is the probability remain the same for a and b (0.8, 0.1, 0.1 for 1%, 5%, and 10%). Is that correct?

Thank you for your help!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi agus;

Welcome!

If you are wrong then I am falling into the same hole.

I do not see how either experiment affects the probability. If I flip a coin twice and get two heads do I now say that the probability of getting a tail has changed? Of course not. I agree with your answer.

Also, supposing we said the probability changed. How am I supposed to compute it without knowing the total number of blowout preventers?

What was the last thing your teacher discussed with you concerning probability?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

hi agus and bobbym,

I, too, seem to be in the same hole. Fortunately, there's a lot more light down here than there is in the 'non-hole' part of the universe.

I suggest we reverse our perspective and call our bit the real world, and the other bit the dark hole!

Even the wording makes no sense to me.

The company makes preventers and some fail. In the population of all preventers the failures can only have one proportion, let's say it's f%. Then that's it, f% fail and (100-f)% work.

Maybe the question is telling us about a series of samples? There's a p= 0.8 probability of getting 1% fail in a sample; a p = 0.1 of getting 5% fail; and a p = 0.1 of getting 10% fail. That tells us about the distribution but it's not anything I have met before.

And we are not told the sample size(s).

Then a single component is sampled and it is a fail. So what? The distribution is unchanged. So the figures remain the same.

If there were only a few components made and we've got a fail then the chances are raised for the rest being good. But we are told nothing about how many are made.

??? The dark is out there. Welcome to the light!

Bob

*Last edited by bob bundy (2011-10-05 00:43:00)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**agus****Member**- Registered: 2011-10-04
- Posts: 2

Hi Bobbym and Bob bun , thanks for your replies, really helpful

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi agus and bob bundy;

Your welcome. Please let me know how this question turns out.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

hi walteroch

Welcome to the forum!

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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