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**Identity****Member**- Registered: 2007-04-18
- Posts: 934

This is a question I missed on the exam:

Sam has 0.8 chance of scoring a goal if he scored his last goal, and 0.7 chance of missing if he didn't score his last goal. Sam scored his first goal. Find the probability that Sam only scores his 2nd goal, given Sam scores at least one of his 2nd, 3rd or 4th goals.

Thanks

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I've never heard of Markov Chains, so this probably isn't the method you're looking for, but:

Same scored his first goal, so the probability that he scores his 2nd is 0.8.

The probability that he then misses his third is 0.2, but the probability that he then misses the fourth is 0.7 (because he missed the previous one).

So the overall probability of (goal, miss, miss) is 0.8*0.2*0.7 = 0.112.

We're told that Sam scores one of his goals, which is the same as saying that he doesn't miss them all.

Using similar reasoning to above, the probability of him missing them all is 0.2*0.7*0.7 = 0.098.

Hence, the probability of him getting one of them is 1-0.098 = 0.902.

Then, the probability of him getting (goal, miss, miss) given that he scores a goal somewhere is 0.112/0.902 ≈ 0.124.

(Or, you could take advantage of the sillily worded question and say that if he scores at least one of his 2nd, 3rd and 4th goals, then he must score his 2nd and so the probability is 1.)

Why did the vector cross the road?

It wanted to be normal.

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**Identity****Member**- Registered: 2007-04-18
- Posts: 934

I thought from the conditional probabiliy:

But 0.112 only gives Pr(Only 2nd).

By the way, Markov Chains are basically another way of describing tree diagrams using matrices. They are really useful (as long as you have a calculator!) because they provide a fully numerical way of solving big conditional probability problems.

Here's an example of how they can be used:

http://en.wikipedia.org/wiki/Examples_of_Markov_chains

*Last edited by Identity (2007-11-19 01:02:06)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

You may be right technically, but since "Only 2nd" is a subset of "At least 1", the intersection of them is the same as just "Only 2nd".

Thanks for that link, by the way. Interesting stuff.

Why did the vector cross the road?

It wanted to be normal.

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**Identity****Member**- Registered: 2007-04-18
- Posts: 934

Yea, I guess you're reasoning is right, although the probability still seems intuitively low.

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