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Hi,
This is one of the past questions from the entrance exam for the graduate school.
I think I know the answer but I would welcome help with the authoritative answer.
Question:
Each time a coin is flipped, it has probability p of a HEAD and q of a TAIL. Let random variable X denote the number of flips until the first HEAD.
(i) Show that P(X>1) = q.
(ii) Write an expression for the conditional probability mass function P(X=x|X>1) in terms of p and q.
What answers did you get (please show your working as well)?Quote:
Originally Posted by itpro
For part i)
(and this is why I am seeking help because I have few pieces together but not a complete
answer - specially not for the part ii)
Total probability of all possible outcomes is 1. We have two equally possible outcomes p and q
and therefore each has probability of 0.5 or 1-p and 1-q.
I recognize P(X>1) as complementary cumulative distribution function ccdf which is equal to 1-cdf
or 1-P(X<=1) or
1- [P(0) + P(1)] = 1 - 0.5 = 0.5 which is q.
Part ii
Conditional probability P(A|B) = [P(A) ^ P(B)] / P(B) = [P(A) * P(B)] / P(B)
In this case I have A = P(X=x) and B=P(X>1)
I have just (attempted to) show that P(X>1) = q
so P(X=x|X>1) = [P(X=x) * q] / q
I can cancel out q so P(X=x|X>1) = P(X=x).
Not sure how much of this is right and how to express P(X=x) as p.
I know that intuitively given that p and q are independent trials each outcome
has 0.5 chance of being a head.
Help please.
(a) is OK.Quote:
Originally Posted by itpro
For (b), do you realise that?
I was "suspecting" at geometric distribution but I did not have much practice with it yet,Quote:
Originally Posted by mr fantastic(a) is OK.
For (b), do you realise that
so I felt it was as good as guessing.
Thank you.