# Help with this conditional PMF question

• Feb 28th 2011, 11:18 AM
itpro
Help with this conditional PMF question
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.
• Feb 28th 2011, 05:33 PM
mr fantastic
Quote:

Originally Posted by itpro
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.

• Feb 28th 2011, 06:39 PM
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.

• Mar 1st 2011, 05:25 PM
mr fantastic
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.

(a) is OK.

For (b), do you realise that $\Pr(X = x) = q^{x-1}p$?
• Mar 1st 2011, 07:24 PM
itpro
Quote:

Originally Posted by mr fantastic
(a) is OK.

For (b), do you realise that $\Pr(X = x) = q^{x-1}p$?

I was "suspecting" at geometric distribution but I did not have much practice with it yet,
so I felt it was as good as guessing.

Thank you.