Hypergeometric Distribution for Sampling Without Replacement; Binomial for Sampling With Replacement
(a) Let $X$ be a hypergeometric random variable counting red balls among 5 balls chosen without replacement from an urn with 5 red and 19 non-red balls.
You seek $P(X = 0) = \frac{{5\choose 0}{19\choose 5}}{{24\choose 5}}=0.2436,$ to four places. Computations in R:
dhyper(0, 5,19, 5)
[1] 0.2735743
prod(19:15)/prod(24:20)
[1] 0.2735743 # your answer
(b) $P(X \ge 1) = 1 - P(X = 0),$ as in the answer (+1) of @lazer-guided-lazerbeam.
(c) Let $Y$ have the distribution $\mathsf{Binom}(n = 5, p=5/24),$ which counts the number of red balls obtained when drawing from the urn with replacement.
You seek $P(Y \ge 1) = 1 - P(Y = 0) = 1- (19/24)^5.$
1 - pbinom(0, 5, 5/24)
[1] 0.6890348
1 - (19/24)^5
[1] 0.6890348
The figure below shows PDFs of both the hypergeometric and the binomial distributions. Notice that the hypergeometric distribution has slightly smaller
variability because fewer choices are available as the number of balls
in the urn decreases.
hdr = "Hypergeometric Probabilities (bars) and Binomial (dots)"
x = 0:5; hpdf = dhyper(x, 5,19, 5); bpdf = dbinom(x, 5, 5/24)
plot(x, hpdf, type="h", lwd=3, col="maroon", xlab="Red Balls", main=hdr)
points(x, bpdf, col="blue", pch=19)
abline(h=0, col="green2"); abline(v=0, col="green2")