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The problem you're working on involves a binomial distribution, as it deals with a fixed number of plants in each greenhouse (25 plants) and the probability that a single plant produces fruit (0.46). Let’s work through each part:

### Part (a)
**Find the probability that exactly 8 plants have fruit on them.**

This can be modeled as a binomial distribution where:
- \( n = 25 \) (number of trials, i.e., number of plants),
- \( p = 0.46 \) (probability of success, i.e., a plant producing fruit),
- \( X = 8 \) (number of successes, i.e., plants with fruit).

The binomial probability mass function is:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
Substituting the values:
\[
P(X = 8) = \binom{25}{8} (0.46)^8 (0.54)^{25-8}
\]

### Part (b)
**Find the probability that more than 12 plants have fruit on them.**

Here, we want to calculate \( P(X > 12) \), which can be expressed as:
\[
P(X > 12) = 1 - P(X \leq 12)
\]
This requires summing the probabilities for all outcomes from 0 to 12:
\[
P(X \leq 12) = P(X = 0) + P(X = 1) + \dots + P(X = 12)
\]
We will calculate this cumulative probability and subtract it from 1 to find the desired value.

### Part (c)
**Find the probability that at least 4 out of 6 greenhouses contain more than 12 plants with fruit.**

This is a binomial distribution problem at a higher level:
- Each greenhouse is considered a trial,
- The probability of more than 12 plants having fruit in a single greenhouse is the answer from part (b), which we denote as \( p_b \),
- We want to find the probability that at least 4 out of 6 greenhouses have more than 12 plants with fruit.

We can model this using a binomial distribution with \( n = 6 \), \( p = p_b \), and \( X \geq 4 \). The required probability is:
\[
P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)
\]

Let me calculate these probabilities for you.Here are the results for each part of the problem:

### Part (a)
The probability that exactly 8 plants have fruit in a greenhouse is approximately **0.0612** (or 6.12%).

### Part (b)
The probability that more than 12 plants have fruit in a greenhouse is approximately **0.3429** (or 34.29%).

### Part (c)
The probability that at least 4 out of 6 greenhouses contain more than 12 plants with fruit is approximately **0.1098** (or 10.98%).

Would you like more details on any of the steps? Here are 5 related questions you might want to explore:

1. What is the probability of exactly 12 plants having fruit?
2. How does changing the probability \(p = 0.46\) affect the outcome for part (b)?
3. What would the probability be if Olivia had 10 greenhouses instead of 6?
4. How can we approximate binomial distributions using normal distributions for large values of \(n\)?
5. How would the answers change if each greenhouse contained 30 plants instead of 25?

**Tip:** The binomial distribution is useful when dealing with a fixed number of trials with two outcomes, like "fruit" or "no fruit" in this case.

Olivia grows orchids. The probability of a particular plant producing fruit is 0.46. Olivia has many greenhouses, each of which holds 25 plants. Find the probability for parts (a), (b), and (c).

Solve a binomial distribution problem involving orchids, where the probability of a plant producing fruit is 0.46. This question covers calculating probabilities for exactly 8 plants, more than 12 plants, and at least 4 greenhouses with more than 12 plants containing fruit. Ideal for high school students in Grades 10-12.

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