This is a problem involving binomial probability, where the probability of a given number of successes in a fixed number of trials can be computed using the binomial formula:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$

Where:

• $n = 25$ is the sample size,
• $p = 0.80$ is the probability of success (i.e., a man would marry the same woman again),
• $k$ is the number of successes (the specific number of men),
• $\binom{n}{k}$ is the binomial coefficient.

1. Probability that exactly 17 would marry the same woman:

We need to compute $P(X = 17)$ for a binomial distribution.

$P(X = 17) = \binom{25}{17} (0.80)^{17} (0.20)^{8}$

Calculating this:

$P(X = 17) = \frac{25!}{17!(25-17)!} (0.80)^{17} (0.20)^8$

Using a calculator or statistical software, this probability is approximately:

$P(X = 17) \approx 0.00499 = 0.499\%$

2. Probability that more than 22 would marry the same woman:

This is the probability $P(X > 22)$, which is equivalent to $P(X = 23) + P(X = 24) + P(X = 25)$.

We calculate these probabilities individually using the binomial formula.

• $P(X = 23) = \binom{25}{23} (0.80)^{23} (0.20)^2$
• $P(X = 24) = \binom{25}{24} (0.80)^{24} (0.20)^1$
• $P(X = 25) = \binom{25}{25} (0.80)^{25} (0.20)^0$

Adding these values together will give us $P(X > 22)$.

Using calculations, the probability $P(X > 22) \approx 0.130$.

3. Probability that at most 4 would marry the same woman:

This is $P(X \leq 4)$, which is the sum of the probabilities for $P(X = 0), P(X = 1), P(X = 2), P(X = 3), P(X = 4)$.

• $P(X = 0) = \binom{25}{0} (0.80)^{0} (0.20)^{25}$
• $P(X = 1) = \binom{25}{1} (0.80)^{1} (0.20)^{24}$
• $P(X = 2) = \binom{25}{2} (0.80)^{2} (0.20)^{23}$
• $P(X = 3) = \binom{25}{3} (0.80)^{3} (0.20)^{22}$
• $P(X = 4) = \binom{25}{4} (0.80)^{4} (0.20)^{21}$

Summing these will give the probability $P(X \leq 4)$.

The calculation gives $P(X \leq 4) \approx 0.0000003$, which is essentially 0%.

4. Probability that at least 20 would marry the same woman:

This is $P(X \geq 20)$, which is the sum of $P(X = 20) + P(X = 21) + P(X = 22) + P(X = 23) + P(X = 24) + P(X = 25)$.

• $P(X = 20) = \binom{25}{20} (0.80)^{20} (0.20)^{5}$
• $P(X = 21) = \binom{25}{21} (0.80)^{21} (0.20)^{4}$
• $P(X = 22) = \binom{25}{22} (0.80)^{22} (0.20)^{3}$
• $P(X = 23) = \binom{25}{23} (0.80)^{23} (0.20)^{2}$
• $P(X = 24) = \binom{25}{24} (0.80)^{24} (0.20)^{1}$
• $P(X = 25) = \binom{25}{25} (0.80)^{25} (0.20)^{0}$

Summing these gives $P(X \geq 20) \approx 0.616$.

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Let me know if you'd like further details on any calculation!

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Follow-up questions:

1. How would the probabilities change if the sample size were increased to 30?
2. How does the binomial distribution differ from other probability distributions like Poisson or normal?
3. What is the difference between cumulative and individual probabilities in binomial distributions?
4. How can these calculations be adjusted for a different probability of success?
5. What is the relevance of the law of large numbers in relation to binomial distributions?

Tip:

In binomial probability, as the number of trials increases and the probability of success is moderate, the distribution tends to look more like a normal distribution.