To solve these problems, we'll use the binomial probability formula, which is given by:

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

where:

• $n$ is the number of trials,
• $k$ is the number of successful trials,
• $p$ is the probability of success on an individual trial,
• $\binom{n}{k}$ is the binomial coefficient.

Given:

• $p = 0.59$ (probability that one American supports increased funding),
• $1 - p = 0.41$ (probability that one American does not support increased funding),
• $n = 4$.

Part (a)

Problem: All 4 of them support increased funding for public schools.

Solution: Here, $k = 4$.

$P(X = 4) = \binom{4}{4} (0.59)^4 (0.41)^0 = 1 \times (0.59)^4 \times 1 = (0.59)^4$

Calculating $(0.59)^4$:

$(0.59)^4 \approx 0.121$

Part (b)

Problem: None of the 4 support increased funding for public schools.

Solution: Here, $k = 0$.

$P(X = 0) = \binom{4}{0} (0.59)^0 (0.41)^4 = 1 \times 1 \times (0.41)^4 = (0.41)^4$

Calculating $(0.41)^4$:

$(0.41)^4 \approx 0.028$

Part (c)

Problem: At least one of the 4 supports increased funding for public schools.

Solution: The probability that at least one supports increased funding is the complement of the probability that none support increased funding.

$P(\text{at least one}) = 1 - P(X = 0)$

Using the result from part (b):

$P(\text{at least one}) = 1 - 0.028 = 0.972$

So, the probabilities are:

• (a) $P(X = 4) \approx 0.121$
• (b) $P(X = 0) \approx 0.028$
• (c) $P(\text{at least one}) \approx 0.972$

Would you like more details on any part of the solution, or do you have any other questions?

Here are 5 questions you might want to ask next:

1. How do you calculate the mean and variance of a binomial distribution?
2. What is the probability of exactly 2 out of 4 Americans supporting increased funding?
3. How does the probability change if the sample size is increased to 10?
4. What is the difference between binomial and normal distributions?
5. How can we use a probability distribution table to solve binomial problems?

Tip: Understanding the complement rule ($P(\text{at least one}) = 1 - P(\text{none})$) can simplify many probability problems.