A Negative binomial problem $P(X \ge 5)$ equals? 
Consider a sequence of independent Bernoulli trials with the
probability of success in each trial being $\dfrac{1}{3}$. Let $X$
denote the number of trials required to get the second success. Then
$P(X \ge 5)$ equals.
$A=\dfrac{3}{7}$
$B=\dfrac{16}{27}$
$C=\dfrac{2}{3}$
$D=\dfrac{9}{13}$

This is a problem of negative binomial $r=2$
$P(X=x)= {x+r-1 \choose r-1}p^rq^x ;\ \ x=0,1,2..$
$P(X\ge5)=1-P(X < 5) \implies 1-(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4))$
This is very time consuming how do I save time on this problem.
How do I utilize CDF of negative binomial distribution in this problem?
${\displaystyle k\mapsto 1-I_{p}(k+1,\,r),}$ the regularized incomplete beta function
 A: Another way to think about the question is to observe that if $X \ge 5$, that means that at least $5$ trials are needed to observe the second success.  This is equivalent to saying that no more than $1$ success occurs in the first four trials.
So there are two cases:  $0$ successes in four trials, or exactly $1$ success in $4$ trials.
In the first case, the probability is $(1-p)^4$.  In the second, the probability is $\binom{4}{1} p (1-p)^3$.  So the total probability is $$(1-p)^4 + 4p(1-p)^3$$ where $p = 1/3$.
A: Easy is $B:\frac{16}{27}$
$$P(X\geq 5)=1-P(X\leq 4)=1-\binom{2-1}{2-1}\Bigg(\frac{1}{3}\Bigg)^2\cdot\Bigg(\frac{2}{3}\Bigg)^0-\binom{3-1}{2-1}\Bigg(\frac{1}{3}\Bigg)^2\cdot\Bigg(\frac{2}{3}\Bigg)^1-\binom{4-1}{2-1}\Bigg(\frac{1}{3}\Bigg)^2\cdot\Bigg(\frac{2}{3}\Bigg)^2=$$
$$=1-\frac{3}{27}-\frac{4}{27}-\frac{4}{27}=\frac{16}{27}$$

Note that there are at least 2 parametrizations of NBinomial. I used the one counting the trials before s successes whose pmf is
$$P(X=n)=\binom{n-1}{s-1}\theta^s(1-\theta)^{n-s}$$
Where the support is $n=s,s+1,s+2,...$
