(Please provide necessary R statements) A utility company might offer electrical rates based on time-of-day consumption to decrease the peak demand in a day. Enough customers need to accept the plan for it to be successful. Suppose that among 50 major customers, 15 would accept the plan. The utility selects 10 major customers randomly (without replacement) to contact and promote the plan. (a) What is the probability that exactly two of the selected major customers accept the plan? (b) What is the probability that at least one of the selected major customers accepts the plan?

This problem can be modeled using **hypergeometric distribution**, as it involves sampling without replacement from a population.

### (a) Probability that exactly two of the selected major customers accept the plan:
Let’s define the variables:
- Population size \( N = 50 \)
- Number of successes in the population (customers who would accept the plan) \( K = 15 \)
- Sample size (number of customers contacted) \( n = 10 \)
- Number of successes in the sample \( k = 2 \)

The probability of selecting exactly \( k \) successes in a hypergeometric distribution is given by:

\[
P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}
\]

In R, you can use the function `dhyper(k, K, N - K, n)` to calculate this probability:

```r
# Parameters
N <- 50   # Total number of major customers
K <- 15   # Number of customers who would accept the plan
n <- 10   # Number of customers contacted
k <- 2    # Number of customers accepting the plan

# Probability that exactly 2 customers accept the plan
prob_2_accept <- dhyper(k, K, N - K, n)
prob_2_accept
```

### (b) Probability that at least one of the selected major customers accepts the plan:
The probability that at least one customer accepts the plan is the complement of the probability that no customer accepts the plan (i.e., \( k = 0 \)).

\[
P(X \geq 1) = 1 - P(X = 0)
\]

In R, this can be computed as:

```r
# Probability that no customer accepts the plan
prob_0_accept <- dhyper(0, K, N - K, n)

# Probability that at least one customer accepts the plan
prob_at_least_1_accept <- 1 - prob_0_accept
prob_at_least_1_accept
```

### Recap:
- Part (a) calculates the probability of exactly 2 customers accepting the plan.
- Part (b) computes the probability that at least 1 customer accepts the plan.

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Do you want more details or have any questions?

Here are 5 related questions:
1. What is the difference between sampling with and without replacement in probability?
2. How does hypergeometric distribution differ from binomial distribution?
3. What if 20 customers accepted the plan instead of 15? How would this affect the probabilities?
4. How would you calculate the cumulative probability of 3 or fewer customers accepting the plan?
5. How would the solution change if the utility selected a different number of customers to contact?

**Tip:** The hypergeometric distribution is ideal for scenarios where the population is finite and each selection influences future selections, unlike the binomial distribution which assumes independence between trials.

A utility company might offer electrical rates based on time-of-day consumption to decrease the peak demand in a day. Enough customers need to accept the plan for it to be successful. Suppose that among 50 major customers, 15 would accept the plan. The utility selects 10 major customers randomly (without replacement) to contact and promote the plan. (a) What is the probability that exactly two of the selected major customers accept the plan? (b) What is the probability that at least one of the selected major customers accepts the plan?

Learn how to solve a probability problem involving hypergeometric distribution. In this example, a utility company randomly selects 10 customers from a pool of 50, where 15 are likely to accept an electricity plan. Calculate the probability of exactly 2 customers accepting the plan and the probability of at least one accepting the plan. Suitable for students in Grades 11-12.

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