I'm pretty sure if you substitute it into the equation then LHS=RHS. I didn't check the convergence of it although I think comparison to a standard geometric series guarantees the convergence(Original post by gff)
Spoiler:Show
Can you elaborate on the convergence of this power series you've posted? I don't see how it works out.
Why this is a reasonable approximation?
Announcements

 Follow
 21
 14042012 23:36

 Follow
 22
 14042012 23:57
(Original post by TheMagicMan)
I'm pretty sure if you substitute it into the equation then LHS=RHS. I didn't check the convergence of it although I think comparison to a standard geometric series guarantees the convergence

 Follow
 23
 15042012 01:30
(Original post by hassi94)
...
(Original post by Slumpy)
This is right.
(Original post by gff)
... 
 Follow
 24
 15042012 01:32
(Original post by Dog4444)
Any hints on how to approximate y in w=lnM+y? I assume y is a random variable.
(In the case in question there's 2/3 chance the coin you have is the biased one.) 
 Follow
 25
 15042012 01:38
(Original post by Slumpy)
The chance of you having the biased coin is not 1/129. If you just picked a coin at random it would be, but if you picked up a coin, flipped it a thousand times, and got heads every time, the odds would be hugely in favour of you having picked the biased coin. Can you see this makes sense?
(In the case in question there's 2/3 chance the coin you have is the biased one.)
There are 129 coins. I pick one and flip it over 9000 times. But the chance that it's biased is still 1/129 no matter what?
And when 2/3 comes from?Last edited by Dog4444; 15042012 at 01:40. 
 Follow
 26
 15042012 01:44
(Original post by Dog4444)
Nope.
There are 129 coins. I pick one and flip it over 9000 times. But the chance that it's biased is still 1/129 no matter what?
I have 2 coins, one is double headed, one is normal.
If I've flipped a coin 3 times and got heads every time. The odds of this happening are:
P(picked an unbiased coin)*P(heads each time)=1/2*(1/2)^3=1/16
P(picked biased coin)*P(heads every time)=1/2*1^3=1/2
The other 7/16's are situations in which you pick an unbiased coin, and don't get 3 heads.
So, with the 3 heads options, the probability that the coin was biased was (1/2)/(1/2 + 1/16)=8/9
Yes? 
 Follow
 27
 15042012 01:56
(Original post by Slumpy)
So, with the 3 heads options, the probability that the coin was biased was (1/2)/(1/2 + 1/16)=8/9
Yes?
And still can't follow the logic. What does other 9/32 mean then (probability the coin is unbaised?) and what (322*9)/32 mean then?Last edited by Dog4444; 15042012 at 01:58. 
 Follow
 28
 15042012 02:00
(Original post by Dog4444)
Can't follow this line. (1/2)/(1/2 + 1/16)=9/32.
( http://www.wolframalpha.com/input/?i...2F2+%2B+1%2F16 if you won't take it on faith)
Editto check; it's division, not multiplication.
(Original post by Dog4444)
But still can't follow the logic.
You pick a coin and flip it once, getting a heads. Can you see the probability that you picked the double heads coin is 1?
(Original post by Dog4444)
What does other 9/32 mean then (probability the coin is unbaised?) and what (322*9)/32 mean then?Last edited by Slumpy; 15042012 at 02:03. 
 Follow
 29
 15042012 02:07
(Original post by Slumpy)
No, what I wrote there is correct.
( http://www.wolframalpha.com/input/?i...2F2+%2B+1%2F16 if you won't take it on faith)
Editto check; it's division, not multiplication.
Ok, consider an even simpler(if slightly odder) case. You have 2 coins. One is heads on both sides, the other is tails on both.
You pick a coin and flip it once, getting a heads. Can you see the probability that you picked the double heads coin is 1?
Edit: Don't really know where you're getting this stuff from tbh.
But still can't follow why you divided instead of multiplying. Can you bring more maths in it like P(AB) and stuff? 
 Follow
 30
 15042012 02:30
(Original post by Dog4444)
Yeah, now I see how crappy my logic is.
But still can't follow why you divided instead of multiplying. Can you bring more maths in it like P(AB) and stuff?
Event A is the event you pick coin X. Event B is the event you throw three heads in a row.
P(A)=1/2, pretty clearly, but we want P(AB)
P(AB)=P(AnB)/P(B)
Now, we know that P(B)=(1/2)*P(BA)+(1/2)*P(BA'), yeah? (This is the law of total probability: http://en.wikipedia.org/wiki/Law_of_total_probability )
so you get P(AB)=2*P(AnB)/(P(BA)+P(BA')).
P(BA)=P(BnA)/P(A)=2P(BnA) and P(BA')=P(BnA')/P(A')=2P(BnA') this just comes from the definitions
So we now have P(AB)=P(AnB)/(P(AnB)+P(A'nB))
Now, P(AnB)=P(A)=1/2, and P(A'nB)=1/2*(1/2)^3=1/16
So we get P(AB)=(1/2)/(1/2+1/16)
Does this make sense?Post rating:1 
 Follow
 31
 15042012 02:41
(Original post by Dog4444)
Any hints on how to approximate y in w=lnM+y? I assume y is a random variable.

 Follow
 32
 15042012 04:42
(Original post by Slumpy)
So, we have the 2 coins, coin X is HH, coin Y is normal.
Event A is the event you pick coin X. Event B is the event you throw three heads in a row.
P(A)=1/2, pretty clearly, but we want P(AB)
P(AB)=P(AnB)/P(B)
Now, we know that P(B)=(1/2)*P(BA)+(1/2)*P(BA'), yeah? (This is the law of total probability: http://en.wikipedia.org/wiki/Law_of_total_probability )
so you get P(AB)=2*P(AnB)/(P(BA)+P(BA')).
P(BA)=P(BnA)/P(A)=2P(BnA) and P(BA')=P(BnA')/P(A')=2P(BnA') this just comes from the definitions
So we now have P(AB)=P(AnB)/(P(AnB)+P(A'nB))
Now, P(AnB)=P(A)=1/2, and P(A'nB)=1/2*(1/2)^3=1/16
So we get P(AB)=(1/2)/(1/2+1/16)
Does this make sense?

 Follow
 33
 15042012 09:47
(Original post by gff)
I may be a bit ignorant, but to deduce the relationship I let , and these power series don't fit into that.

 Follow
 34
 15042012 10:19
(Original post by hassi94)
I don't think so.
I think we'd say:
P(coin is unbiased) = 128/129
P(coin is biased) = 1/129
P(unbiased coin is heads 8 times) = (1/2)^8 = 1/256
P(biased coin is heads 8 times) = 1
So the probability of an unbiased coin doing what has been done is 1/258
And the probability of a biased coin doing what is done is 2/258
So the probability we have a biased coin is 2/3 and the probability we have a unbiased coin is 1/3
Then finally the probability it will be heads again is (2/3)(1) + (1/3)(1/2) = 5/6
I don't really know if this is correct, and sorry I couldn't use proper notation  I've never really done probability except a tiny bit in S1 so I've just done this out of logic alone.
Also could you explain the underlined if you can 
 Follow
 35
 15042012 10:44
is question 4, 1600 by any chance?
6000 numbers
remove multiples of 2 (half of them)
60003000=3000
remove multiples of 3 (half of which are multiples of 2)
3000(2000/2)=2000
remove multiples of 5 ( removing multiples of 2 and 3, there are 4 between 560, therefore 400 up to 6000)
2000400=1600 
 Follow
 36
 15042012 11:11
(Original post by Dmon1Unlimited)
Sorry, I didn't understand the bold bit, how did the 6 change to an 8?
Also could you explain the underlined if you can
By "So the probability of an unbiased coin doing what has been done is 1/258" I meant the probability of an unbiased coin having got heads 8 times is (128/129)*(1/256) = 1/258
As for the underlined bit:
We have a situation in which we have gotten heads 8 times. The probability of this happening is 1/258 for the unbiased coin (as shown above) and 2/258 for the biased coin. Therefore, it is twice as likely that the coin being talked about in the question is a biased coin. So then we can say the probability that the coin the question is referring to is a biased coin is 2/3, and 1/3 for unbiased.
Sorry I'm not great at explaining  I'm not very good at probability 
 Follow
 37
 15042012 11:24
(Original post by Dmon1Unlimited)
is question 4, 1600 by any chance?
6000 numbers
remove multiples of 2 (half of them)
60003000=3000
remove multiples of 3 (half of which are multiples of 2)
3000(2000/2)=2000
remove multiples of 5 ( removing multiples of 2 and 3, there are 4 between 560, therefore 400 up to 6000)
2000400=1600
I did it in a pretty roundabout way (from now on I'll write number of multiples of x as M(x)):
M(2) = 3000
M(3) = 2000
M(5) = 1200
If we take away all of these from 6000 we are also taking 2 groups of M(6), M(10) and M(15)  so we need to add these back in  however this will add in an extra M(30) so we must take that away.
M(6) = 1000
M(10) = 600
M(15) = 400
M(30) = 200
So the answer is 6000  M(2)  M(3)  M(5) + M(6) + M(10) + M(15)  M(30)= 6000  3000  2000  1200 + 1000 + 600 + 400  200 = 1600
Interview stress made me overcomplicate things I think haha 
 Follow
 38
 15042012 13:02
(Original post by hassi94)
Sounds right, I got this question in interview
I did it in a pretty roundabout way (from now on I'll write number of multiples of x as M(x)):
M(2) = 3000
M(3) = 2000
M(5) = 1200
If we take away all of these from 6000 we are also taking 2 groups of M(6), M(10) and M(15)  so we need to add these back in  however this will add in an extra M(30) so we must take that away.
M(6) = 1000
M(10) = 600
M(15) = 400
M(30) = 200
So the answer is 6000  M(2)  M(3)  M(5) + M(6) + M(10) + M(15)  M(30)= 6000  3000  2000  1200 + 1000 + 600 + 400  200 = 1600
Interview stress made me overcomplicate things I think haha 
 Follow
 39
 15042012 13:11
(Original post by TheMagicMan)
This is basically inclusion exclusion. Anyway a slightly different method is to consider the numbers up to thirty and multiply by 200 
 Follow
 40
 15042012 14:13
(Original post by hassi94)
Never heard of inclusion exclusion Yeah I think that is the best way to do it (the way you said).
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: April 15, 2012
Share this discussion:
Tweet
Related discussions:
 Taylor and Maclaurin Series
 binomial expansion
 Basic Stat True/False questions..
 Proton Decoupled NMR
 Total energy of gamma rays
 Ocr mei S2 6th june 2013
 Hard Mechanics Discussion Thread
 Help with this log graph problem
 Can someone check these practice continuity questions ...
 Girls: Do you view men as the weaker sex?
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.