Question

Suppose $X$ is a random variable which takes values $0,1,2,3,...$ and $P(X=r)=p{q}^r$ where $0<p<1,q=1-p$ and $r=0,1,2,...$, Then

• A. $P(X\ge n)=q^n$
• B. $P(X\ge m+n|X\ge m+n|X\ge m)=P(X\ge n)$
• C. $P(X=m+n|X\ge m+n|X\ge m)=P(X=n)$
• D. None of these

Answer 1

Answer: (A), (B), (C)

$P(X\ge n)=q^n$
$P(X\ge m+n|X\ge m+n|X\ge m)=P(X\ge n)$
$P(X=m+n|X\ge m+n|X\ge m)=P(X=n)$

$P(X\ge n)=\sum _{k=n}^{\infty }P(X=k)=\frac{p{q}^n}{1-q}=q^n$
$P(X\ge m+n|X\ge m)=\frac{P(X\ge m+n)}{P(X\ge m)}=\frac{q^{m+n}}{q^m}=q^n$

$P(X=m+n|X\ge m)=\frac{P(X=m+n)}{P(X\ge m)}=p{q}^n$.