Normal approximation to negative binomial (Sheldon Ross)

I am stuck at the following question -

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?

My attempt, in the easiest way, is to derive the probability using negative binomial distribution, which comes as-

required probability $$= \displaystyle\sum_{i=100}^{\infty} \binom{i}{49}0.4^{50}0.6^{(i-49)} \approx 0.973$$

However because of the summation to infinity, the book suggests that I have to use Normal approximation here and I don't have any idea on how to proceed.

• Instead, calculate the mean and standard deviation of the negative binomial variate, and use those as the parameters of an approximating Normal distribution. I should note that some of Ross's problems were probably written decades ago; summing until you reach the 99.999th percentile of the NB distribution will certainly give you a reasonable approximation, but back in the day was not a generally feasible strategy. Commented Oct 22, 2021 at 14:44
• Because the summation beginning with $i=49$ must equal $1,$ you can obtain the answer with a relatively short finite sum. Indeed, the first many terms of this finite sum are so small you can safely neglect them; a good approximation can be obtained by summing from $i=80$ through $i=99,$ for instance. Here is a one-line R calculation using full double precision. Use it to check your approximation. with(data.frame(i=50:100-1), 1 - sum(exp(lchoose(i, 50-1) + 50*log(0.4) + (i-50+1)*log(1-0.4))))
– whuber
Commented Oct 22, 2021 at 17:07

So here you would have a mean of $$\frac{50}{0.4} = 125$$ and variance of $$\frac{0.6\times 50}{0.4^2} = 187.5$$ so standard deviation of about $$13.693$$ and would calculate $$1-\Phi\left(\frac{100.5 - 125}{13.693}\right) \approx 0.9632$$
An alternative approach would be to transform the question into a binomial distribution as ask for the probability that after $$100$$ attempts you have $$49$$ or fewer successes. This gives a direct calculation of $$\sum\limits_{j=0}^{49} \binom{100}{j}0.4^{j}0.6^{(100-j)} \approx 0.9729$$ the same as yours.
Now for the normal approximation to the binomial you would have a mean of $$100 \times 0.4 = 40$$ and variance of $$100 \times 0.4 \times 0.6 = 24$$ so standard deviation of about $$4.899$$ and would calculate $$\Phi\left(\frac{49.5 - 40}{4.899}\right) \approx 0.9738$$ which is a bit closer
• @AnkitSeth No - I mean exactly $100$ attempts. If you imagine an unlimited number of attempts, then having the $50$th success to be later than the $100$th attempt is equivalent to having $49$ or fewer successes in the first $100$ attempts. Commented Oct 23, 2021 at 15:34