Re: Probability with dice..
- From: "mina_world" <mina_world@xxxxxxxxxxx>
- Date: Tue, 30 Oct 2007 15:45:18 +0900
"mina_world" <mina_world@xxxxxxxxxxx> wrote in message
news:fg6g08$su1$1@xxxxxxxxxxxxxxxxxxx
Hello sir~
When I throw the dice,
If even , then I win.
If odd, then I throw again.
(Namely, If even, then END.)
I can throw "6 times" at most.
Let X be a random variable with the number of times I throw the dice.
Find P(X=2) + P(X=4).
-----------------------------------------------------
If I can throw "Infinitely many", easy.
P(X = 2) = (1/2)(1/2) = 1/4 with odd and even.
P(X = 4) = (1/2)^4 = 1/16 with odd and odd and odd and even.
so, P(X=2) + P(X=4) = 5/16.
But I can throw "6 times" at most.
so, P(X=1) + ....+ P(X=6) = (1/2) + ... + (1/2)^6 = 63/64.
so, answer is (5/16) / (63/64).
is this possible ?
But P(S = all events) = 63/64 =/= 1.
so, there does not exist the p.d.f of X.
is this unnecessary ?
Oh, very sorry.
I repair my problem.
When I throw a die,
If even , then I win.
If odd, then I throw again.
(Namely, If even, then END.)
I can throw "6 times" at most.
Let X be a random variable
with the number of times I throw a die "UNTIL I WIN".
Find P(X=2) + P(X=4).
--------------------------------------
I add "until I win". and "dice ==> die".
and then, P(X=1) + ....+ P(X=6) = (1/2) + ... + (1/2)^6 = 63/64.
This is true.
and
But P(S = all events) = 63/64 =/= 1.
so, there does not exist the p.d.f of X.
How do you think about it ?
.
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