Every day Jo practices her tennis serve by continually serving until
she has had a total of $50$ successful serves. If each of her serves
is, independently of previous ones, successful with probability 0.4,
approximately what is the probability that she will need more than 100
serves to accomplish her goal?
Attempt
Let $X$ be number of serves (trials) until 50 successful ones (success). Then, $X$ is negative binomial with parameters $r=50$ and $p = 0.4$. and thus,
$$ P(X \geq 100 ) = \sum_{i=100}^{\infty} {i-1 \choose 49} 0.6^{i-50} 0.4^{50} $$
Now, calculating this would be hard to put in calculator. The question asks for approximate probability, hence this means I may have to approximate using Normal approximation but this only works for Binomial rv and here we have negative binomial. Am I misinterpreting the problem?
Best Answer
On average, how many serves does she need to get one successful serve?
The probability of success on each serve is $0.4= \dfrac 2 5,$ so the average number of serves to get one success is $\dfrac 5 2 = 2.5.$ Check out the expected value of the geometric distribution, which is the distribution of the number of trials needed to get one success. (Do not confuse this with the other geometric distribution, which is the number of trials before the first success, which may be $0.$)
So to get $50$ successes on average requires $50$ times that many attempts.
A sum of $50$ independent geometrically distributed random variables with the same mean has $50$ times the mean and $50$ times the variance, thus $\sqrt{50}$ times the standard deviation, as does one of the geometrically distributed random variables. Thus the central limit theorem does apply.