Recent content by themaestro

Just wondering if anyone has any ideas about where I have gone wrong? i reckon there must be some other relationship between P(Af) and P(Apo) that I have not taken account of.

Dont think the equations are vaild as infinity is not a number so can not have e.g A=infinity. You can only say A tends to infinity. (and here A can not be a number because a number is fixed and can not tend. A must be an expression, e.g the sum of a infinite series, if for example y=1/x, the...

here is details of my attempt. P(Af)=P(Af| (A and B)) *P(A and B) + P(Af| not(A and B))*P(not(A and B)) (1) since P(Af| not(A and B)) can only occur if Team A reaches 6 points and B does not then this is equal to (P(A)- P(A and B)) / (P(not(A and B)) sub into (1) and...

is this problem not related to 'Benford’s Law' ?

If I understand problem correctly then P(S[2]=i)=P(S[2]=i|S[1]\neqi)*P(S[1]\neqi) so for i=1 then P(S[2]=1)=(1/8)*(4/5) will be the same method for other elements

I would have thought he would be after P(Bitten by own pet|Bitten) which would be 87/95

Was hoping someone might be able to assist in this problem Scenario is have a American football game with two team A and team B. Am trying to work out the probability of each team getting to 6 points first, from the following infomation: Let A be the event team A scores equal to or more...

thanks techmologist, will test this to see what results it gives. The logic looks to be sound as far as I can tell

this is what I do not think is right, I don't know what the answer should be in this example but looking at other examples. If X=2 and Y=3 then this equation gives (X/X+Y)^3=(2/5)^3=0.008. but infact the answer should be 0 as it is impossible for there to be 3 reds in a row if there are only...

I think these are the numbers your after 0.25^900= 1.4*10^(-542) and (0.25^(900))^23000= 1.5* 10^(-12462642) didnt need a cray computer; it is not actually a difficult calculation, your calculator probably can't deal with such small numbers! hope...

Hi Was hoping you could clear this thing up about the prior probability (X/X+Y)^3 and exactly what you mean by this. One thing I know its not is the initital probability of the first 3 balls being red as this is (X')(X'-1)(X'-2)/[(X'+Y')(X'+Y'-1)(X'+Y'-2)] also it can not be the initital...

would it not be (X')(X'-1)(X'-2)/[(X'+Y')(X'+Y'-1)(X'+Y'-2)] instead of (X/[X+Y])^3 as the number of remaining balls reduces by one after every pick? Regarding your main algorithm i think it looks fine, my only issues would be with actually using this as wouldn't there be a very large...

Thanks for taking the time to have a look at this. I hadnt considered looking at in this way before. One problem is that the 3 reds can occur anywhere in the sequence, so by your method would we need to sum for all combinations of n(x)n(y)? i.e would it be Double sum(...

Hi This question has been driving me mad, hoping someone will be able to help me The question is: A bag contains X Red balls and Y Blue Balls. A ball is picked from the bag (at random) one at a time until all the balls have been picked. What is the Probability that 3 successive red balls...