$Y=X_1+ X_2+ X_3$, then $P(Y\geq 5)$

Let $X_1, X_2, X_3$ be i.i.d discrete random variables with p.m.f

$$p(k)=(\frac{2}{3})^{k-1}(\frac{1}{3}),$$

for $k=1, 2, ......$. Let $Y=X_1+ X_2+ X_3$, then $P(Y\geq 5)$?

My work: I tried to find the distribution of $Y$, but I can't. How can I solve this?

Notice that $Y \ge 3$, hence $P(Y \ge 5)=1-P(Y=3)-P(Y=4)$.

There is exactly one way to get $Y=3$, that is $(X_1,X_2,X_3)=(1,1,1)$.

Ways to get $Y=4$ are $(2,1,1),(1,2,1),(1,1,2)$.

Hence compute $1-[p(1)^3 + 3p(1)^2p(2)]$.

From the pdf, $X_1$, $X_2$, and $X_3$ can each be considered the count of trials until the first success in an indefinite sequence of independent Bernoulli trials with identical success rate, $1/3$.   Notice that $(2/3)^{k-1}$ is the probability for $k-1$ consecutive 'failures', $1/3$ the probability for one 'success' in such a system, producing:$$p(k)=(2/3)^{k-1}(1/3)$$

As the sum of three independent counts, $Y$ would then the count of trials until the third success in an indefinite sequence of independent Bernoulli trials with identical success rate, $1/3$.

Under this interpretation, $\mathsf P(Y\geq 5)$ would then be the probability that the third success does not occur before the fifth trial.

• So, is it essentially Negative Binomial and can be represented for example like 10010...1 (1-success, 0-failure), with the number of trials $n \ge 5$, number of successes r=3? – Sargis Iskandaryan Jan 23 '18 at 22:33
• Yes, indeed it is. – Graham Kemp Jan 23 '18 at 23:41