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An urn contains 5 red balls, 7 green balls and 12 yellow balls. You randomly select 5 balls without replacement.

a. What is the probability that there are no red balls among the ones you picked?

b. What is the probability that at least 1 red ball is picked?

c. Repeat problem 2, but you select 5 balls with replacement.

For the first one is it correct to write $(19/24) \times (18/23) \times (17/22)\times (16/21)\times (15/20)$ ?

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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.

enter image description here

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")
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You are correct for a. That is the probability of choosing no red balls.

For b, you want the probability of choosing 1 or more red balls. Hint: Any probability that is not "no red balls" (question a) is part of "1 or more red balls". It is the opposite (compliment) of a.

For c, you will first need to calculate the probability of question a with replacement. Then find the probability of 1 or more red balls with replacement (the compliment of the a with replacement question).

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