Statistical probility, bayes' theorem

Hi,

I am looking for a solution to the following problem, which according to the question can be solved using bayes' theorem:

The same event occurs a 1000 times

Every time, the chance of it being positive is 17%

Everytime 11 events in a row are negative, the 12th is positive with a 100% chance.

There are no external events apart from the aforementioned rules

What is the expected amount of positive events, and how did you reach this conclusion?

My approach was invalid because it was to simplified, yet the end result was close to the actual answer (which still hasn't been provided)

(1000-1000/12)*0.17+1000/12=239.17

That answer is incorrect.

Does anyone have a solution?

Edit: recieved a hint: the chance that the event will be negative 11 times in a row is 12.69%. This implies that that number has something to do with it, otherwise the tip would not have been given

Re: Statistical probility, baye's theorem

Quote:

Originally Posted by

**brentvos** The same event occurs a 1000 times

Every time, the chance of it being positive is 17%

Every 12th event will be positive regardless of chance.

There are no external events apart from the aforementioned rules

That is simply a series of statements.

What is the question?

Re: Statistical probility, baye's theorem

The question is:

What's the amount of positive events given the conditions, and how did you come to that conclusion.

Thanks for the help!

Re: Statistical probility, baye's theorem

Hello, brentvos!

Quote:

I am looking for a solution to the following problem which,

according to the question, can be solved using Bayes' Theorem.

I don't understand the hint; this is not conditonal probability.

The same event occurs a 1000 times

Every time, the chance of it being positive is 17%

Every 12th event will be positive regardless of chance.

There are no external events apart from the aforementioned rules

Find the expected number of positive events.

My approach was invalid because it was too simplified, . How do you know this?

yet the end result was close to the actual answer . And how do you know this

. . (which still hasn't been provided). . . . . . . . . . . if you don't know the answer?

Every 12th event will be positive. .It will be positive.$\displaystyle \left[\frac{1000}{12}\right] = 83$ times.

For the other 917 events, it will be positive 17% of the time.

. . $\displaystyle 0.17 \times 917 \,=\,155.89$ expected positive events.

Therefore, the expectation is: .$\displaystyle 83 + 155.89 \:=\:238.89$ positive events.

Re: Statistical probility, baye's theorem

The question giver told us it had something to do with Bayes' theorem. I didn't understand the hint either, but hoped one of you would.

As for your answer, except for different decimals, I had the same answer as you did. This was incorrect, as told by the question giver.

He gave another tip, saying that the possibility of not having a positive event 11 times in a row is 12.9% thereby declaring that it has something to do with the given statement.

I'll edit the original post to include this

Re: Statistical probility, baye's theorem

Quote:

Originally Posted by

**brentvos** The question giver told us it had something to do with Bayes' theorem. I didn't understand the hint either, but hoped one of you would.

As for your answer, except for different decimals, I had the same answer as you did. This was incorrect, as told by the question giver.

He gave another tip, saying that the possibility of not having a positive event 11 times in a row is 12.9% thereby declaring that it has something to do with the given statement.

Well I agree with Soroban's answer.

Perhaps we all are misreading the intended question.

But rather I suspect that this a case in which the author has a very specific method in mind. So for someone not privy to his/her style any effort at a solution would simply guessing.

Re: Statistical probility, baye's theorem

Ah I've found the mistake.

I copied it wrong.

I changed OP, but the change is:

Every time 11 events in a row have been resulted negative, the 12th is positive.

(the first 11 are negative, the 12th therefore is positive, regardless of any previous chance)

Sorry for the mistake, would you be so kind to look at it again?

Re: Statistical probility, bayes' theorem

Quote:

Originally Posted by

**brentvos** Hi,

I am looking for a solution to the following problem, which according to the question can be solved using bayes' theorem:

The same event occurs a 1000 times

Every time, the chance of it being positive is 17%

Everytime 11 events in a row are negative, the 12th is positive with a 100% chance.

There are no external events apart from the aforementioned rules

What is the expected amount of positive events, and how did you reach this conclusion?

My approach was invalid because it was to simplified, yet the end result was close to the actual answer (which still hasn't been provided)

(1000-1000/12)*0.17+1000/12=239.17

That answer is incorrect.

Does anyone have a solution?

Edit: recieved a hint: the chance that the event will be negative 11 times in a row is 12.69%. This implies that that number has something to do with it, otherwise the tip would not have been given

Hi Brentvos,

The probability that the ith event is positive is 0.17 for i = 1, 2, 3, ..., 11.

For i = 12, 13, 14, ..., 100, the event will be positive with probability 1, if the preceding 11 events are all negative; this happens with probability 0.83^11. If at least one of the 11 preceding events is positive, which happens with probability 1 - 0.83^11, the ith event will be positive with probability 0.17. So the total probability of a positive event is

0.83^11 + (1 - 0.83^11) * 0.17.

Therefore the expected number of positive events is

$\displaystyle 11 \times 0.17 + 989 \times [0.83^{11} + (1 - 0.83^{11}) * 0.17]$