This problem can be approached using the binomial distribution. The binomial distribution applies when there are two outcomes (success or failure), and each trial is independent. Here:

• The probability of a plant belonging to the particular species (success) is $p = 0.09$.
• The number of trials (the sample size) is $n = 9$.

Let $X$ represent the number of plants in the sample that belong to the particular species. Then $X$ follows a binomial distribution:

$X \sim \text{Binomial}(n = 9, p = 0.09)$

The probability mass function of a binomial distribution is:

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

where $k$ is the number of successes, and $\binom{n}{k}$ is the binomial coefficient.

Part (a) - Probability that none of the plants in the sample will be of that particular species:

This corresponds to $P(X = 0)$.

$P(X = 0) = \binom{9}{0} (0.09)^0 (1 - 0.09)^9 = (0.91)^9$

Using this formula, we calculate:

$P(X = 0) \approx (0.91)^9 \approx 0.4228$

So, the probability that none of the plants in the sample belong to the species is approximately 0.4228.

Part (b) - Probability that fewer than 6 of the plants in the sample will be of that particular species:

This corresponds to $P(X < 6)$, or the cumulative probability for $X = 0, 1, 2, 3, 4, 5$.

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

Using the binomial probability formula for each value of $k$:

$P(X = k) = \binom{9}{k} (0.09)^k (0.91)^{9 - k}$

We sum the probabilities for $X = 0$ through $X = 5$.

$P(X < 6) \approx P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) \approx 0.9999$

So, the probability that fewer than 6 of the plants in the sample belong to the species is approximately 0.9999.

Part (c) - Probability that more than 3 of the plants in the sample will be of that particular species:

This corresponds to $P(X > 3)$, or $1 - P(X \leq 3)$.

We calculate $P(X \leq 3)$:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

Summing the individual probabilities for $X = 0, 1, 2, 3$:

$P(X \leq 3) \approx 0.4228 + 0.3741 + 0.1514 + 0.0425 = 0.9908$

Thus, the probability that more than 3 plants belong to the species is:

$P(X > 3) = 1 - P(X \leq 3) = 1 - 0.9908 \approx 0.0092$

So, the probability that more than 3 of the plants in the sample belong to the species is approximately 0.0092.

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Would you like more details or clarifications?

Here are some related questions:

1. How do you compute binomial probabilities for a different sample size?
2. What happens to these probabilities if the percentage of the species changes?
3. How does the cumulative distribution function (CDF) help in these types of problems?
4. Can you derive the mean and variance of a binomial distribution?
5. How does the binomial distribution relate to the normal distribution in large samples?

Tip: When dealing with binomial problems involving a small probability and a large number of trials, the Poisson approximation can sometimes simplify the calculations.