The problem states that you play a weekly lottery that has a $1\%$ chance of winning each week, independently. Let $X$ represent the number of weeks you do not win the lottery before your first win. What is the probability that you wait more than one year to win the lottery?

First I'm having some difficulty understanding how the negative binomial works. I think I'm counting $X$ weeks that are failures before my first $k$ success. Is this the correct PMF $f(x)=\binom{x+k-1}{x}p(1-p)^x$, $x=0,1,2,3...$?

and $P(X>51)=1-P(X\leq51) =1-\sum_{x=0}^{51}\binom{x+1-1}{x}(0.01)(1-0.01)^x$?

Any help would be greatly appreciated, thanks


You're counting the number of losses before your first $(k=1)$ win. So the pmf you wrote is correct, and since $k=1$ it reduces to the geometric distribution: \begin{align*} \Pr(\text{lose }n\text{ weeks before first win})= .99^n .01 \end{align*} The expression you wrote for $\Pr(X>51)$ is also correct but it might be more straight forward to write as \begin{align*} \Pr(X\geq 52)&= \sum_{n\geq 52}.99^n.01=.99^{52}*.01*\frac{1}{1-.99} \end{align*} to evaluate it as a geometric series. (In your expression, $\binom{x+1-1}{x}=1$.)


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