probability distribution with discrete random variable

Question

I have tried it out, but finding it difficult to post them in proper format. Please help me with the solution.

Answer 1

(a) Here $X$ is the number of trials until the first "success," where the probability of success on any trial is $p$, and the trials are independent.

We have $X\gt n$ if and only if the first $n$ trials result in $n$ failures. The probability that on any trial we have failure is $1-p$. So the probability of $n$ failures in a row is $(1-p)^n$.

(b) Let $B$ be the event $X\gt n+k$, and let $A$ be the event $X\gt n$. By the definition of conditional probability, we have $$\Pr(B|A)=\frac{\Pr(A\cap B)}{\Pr(A)}.\tag{1}$$ In our case, $A\cap B$ is just the event $X\gt n+k$.

By the calculation in (a), we have $\Pr(X\gt n+k)=(1-p)^{n+k}$, and $\Pr(X\gt n)=(1-p)^n$. Thus, substituting in (1), we get $$\Pr(B|A)=\frac{(1-p)^{n+k}}{(1-p)^n}=(1-p)^k.$$ But by (a), $(1-p)^k=\Pr(X\gt k)$. This completes the proof.

(c) You will need to explain this yourself, in language that is natural to you. We give a brief guide.

We are flipping a coin that has probability $p$ of landing heads, where $0\lt p\le 1$. The random variable $X$ is the number of flips until the first head.

Suppose we have flipped $n$ times without success. The probability that we will need a total of more than $n+k$ flips is the same as the probability that we will need at least $k$ flips if we start a fresh sequence of flips. That's because the coin does not remember that it let you down $n$ times in a row.