This problem is a binomial probability problem because we are dealing with a fixed number of trials (8 freshmen), and each freshman has two possible outcomes: they either agree that same-sex couples should have the right to legal marital status ("yes") or they don't ("no"). The success probability $p$ for a single freshman responding "yes" is 0.719, based on the survey results. We're interested in finding the probability that at least two of the freshmen say "yes."

Step 1: Define the Binomial Distribution

The probability of $k$ successes (freshmen saying "yes") out of $n$ trials (8 freshmen) in a binomial distribution is given by the formula:

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

Where:

• $n = 8$ (the number of freshmen selected),
• $p = 0.719$ (the probability that a freshman says "yes"),
• $k$ is the number of freshmen saying "yes",
• $\binom{n}{k}$ is the binomial coefficient, also called "n choose k."

Step 2: Calculate the Probability of "At Least 2" Freshmen Saying "Yes"

The probability of at least two freshmen responding "yes" is the complement of the probability of fewer than two freshmen saying "yes." This means we need to find the probability of zero and one freshman responding "yes," and subtract that from 1.

$P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$

Now, let's calculate $P(X = 0)$ and $P(X = 1)$.

Step 3: Calculate $P(X = 0)$ and $P(X = 1)$

For $P(X = 0)$:

$P(X = 0) = \binom{8}{0} (0.719)^0 (1 - 0.719)^8 = 1 \times (0.281)^8$

For $P(X = 1)$:

$P(X = 1) = \binom{8}{1} (0.719)^1 (1 - 0.719)^7 = 8 \times 0.719 \times (0.281)^7$

Now, I'll compute these probabilities. Given values p = 0.719 q = 1 - p n = 8

Calculate P(X = 0)

P_X_0 = (q ** 8)

Calculate P(X = 1)

P_X_1 = 8 * p * (q ** 7)

Calculate P(X >= 2)

P_X_at_least_2 = 1 - P_X_0 - P_X_1

P_X_at_least_2, P_X_0, P_X_1The probability that at least two freshmen say "yes" is approximately:

$P(X \geq 2) = 0.9992$

Where:

• The probability of exactly 0 freshmen saying "yes" is approximately $0.00004$,
• The probability of exactly 1 freshman saying "yes" is approximately $0.00080$.

Thus, the probability that at least two freshmen agree is approximately 0.9992.

Let me know if you need more details or have any further questions!

Related Questions:

1. What is the probability that exactly 3 freshmen say "yes"?
2. How would the probability change if the sample size were 10 instead of 8?
3. What does the binomial distribution represent in this context?
4. How does the probability of "at least two" compare to the probability of "at least one"?
5. What would happen if the probability of success was lower than 0.719?

Tip:

For binomial distributions, remember that the complement rule is a powerful tool. When asked for probabilities of "at least" a certain number, subtract the cumulative probability of fewer successes from 1.