3
$\begingroup$

Drought length is referred to as the number of consecutive time intervals in which the water supply remains below a critical value. Consider the drought length as a random variable, denoted as $Y$, which is assumed to have a geometric distribution with $p = 0.409$

  1. What is the probability that a drought lasts exactly $3$ intervals? $(0.0844)$
  2. What is the probability that a drought lasts at most 3 intervals? $(0.878)$

My try:

  1. $(0.409)^3 = 0.06841$.

  2. $(0.409)^3 + (0.409)^2 +(0.409)^1 +(0.409)^0$.

$\endgroup$
6
  • $\begingroup$ What are those numbers to the left of the questions? Are those the answers? $\endgroup$
    – Em.
    Feb 17, 2016 at 13:00
  • $\begingroup$ Yes,they are the given answers $\endgroup$
    – max
    Feb 17, 2016 at 13:00
  • $\begingroup$ Why? Can you explain a little bit about your thoughts? $\endgroup$
    – max
    Feb 17, 2016 at 13:03
  • $\begingroup$ @probablyme the given answers are correct if $p$ actually denotes the probability that the drought ends in the next period. That is, $(1-p)$ represents the probability that it extends. $\endgroup$
    – lulu
    Feb 17, 2016 at 13:08
  • 1
    $\begingroup$ I think the question should be clarified. proablyme and I both read it to suggest that $p$ was the probability that the drought extended (also in the worked solutions it appears that you are also reading in that way, though your algebra is still wrong). if instead you assume that $p$ is the probability that it ends, then the probability that the drought lasts exactly $3$ periods is $(1-p)^3\times p = (.591)^3\times .409\sim .0844$. $\endgroup$
    – lulu
    Feb 17, 2016 at 13:12

1 Answer 1

1
$\begingroup$

The geometric distribution is defined as $$P(Y=k)=p(1-p)^{k-1},\quad k\geq 1$$ Equivalently: $$P(Y>k)=(1-p)^k$$

The first question just asks $P(Y=3)$, and the second question asks $P(Y\leq 3)$, which we can write $1-P(Y>3)$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .