consider the sequence 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
It hasn't had the same number 2 success times but goes much longer than what you have listed as the max (10)
Hi i have this problem, where a fair die is rolled until the same number occurs 2 successive times. Let X be the trial on which the repeat occurs, X=10 find f(x)=p(X=x)
For example, 1,2,3,4,5,6,1,5,3,3
I'm thinking the answer is as follows:
f(x)=P(X=x)=1/n for i=1,2,….,n
Ans: X=10=n=1/10
am i correct or absolutely incorrect heh ?
consider the sequence 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
It hasn't had the same number 2 success times but goes much longer than what you have listed as the max (10)
I don't have emotions and sometimes that makes me very sad.
Dason, why do you speak in code ? heh im not following you
There is no code there. It was a simple example showing you that you "solution" couldn't be correct.
I don't have emotions and sometimes that makes me very sad.
The question states that the expiriment doesnt surpass 10 rolls and that the repeat happens on the 10th roll so how am i wrong ?
P(x=1) = 0 (You can't roll the same number twice in one roll)
P(x=2) = 1/6
P(x=3) = 5/6 * 1/6
p(x=4) = (5/6)^2 * 1/6
So
p(X=5)=(5/6)^3*1/6
p(X=6)=(5/6)^4*1/6
And so on ?
So basically I employ formula:
P(x=m)=q^(m-1)*p ?
Do you understand the rationale behind? How do asterisk come up with that answer?
If yes you should not in doubt.
I dont understand what you mean "if yes you should not in doubt" but the number after the asterisk is the probability of rolling any number on a die, 1/6 since there are 6 numbers on a die.
My question is why are we using 5/6 ? this is the probability that any OTHER different number is rolled right
Last edited by ITman95; 11-21-2013 at 08:58 AM.
1/6 is the probability of matching the previous roll. (stop rolling)
5/6 is the probability of not matching the previous roll. (continue rolling)
This is basically a negative binomial distribution, except on the first roll probability of stopping = 0 and the probability of continuing is 1.
where
r= number of failures = 1
k {1,2,3 ... } = number of successes
p = probability of success = 5/6
(1 - p) = probability of failure = 1/6
By convention we usually call the stopping event (rolling the same number twice in a row) a failure. It may seem more natural to call this a success, but it just means the stopping event.
Because r=1 the first term simplifies to = = 1
So so for P(X = 10) we need
SSSSSSSSSF =
ITman95 (11-21-2013)
I see, thank you for clarifying.
I think you made a mistake wouldnt p(x=1)=1/6 and p(x=2)=(5/6)*(1/6) and p(x=3)=(5/6)^2 * (1/6) and so on so that its P(X = x) = (5/6)(x-1) × (1/6) or is this just in the case where if you are looking to roll a specific number for example a 5 instead of a repeat which is whats asked in the question ?
Last edited by ITman95; 11-21-2013 at 04:30 PM.
I don't see how p(x=1) could equal 1/6.
How could you roll the same number twice on the first roll? Unless you are not counting the first roll... then it is just semantics.
alright i was just confusing my self with another problem thats all. Thank you for your help asterisk.
Tweet |