To solve this, we will use the binomial probability formula since each M&M selection can be viewed as a Bernoulli trial (it is either green or not green), and we are interested in the probability of getting a certain number of green M&M's.

The binomial probability formula is:

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

Where:

• $P(X = k)$ is the probability of getting exactly $k$ successes (green M&M's in this case) in $n$ trials.
• $n$ is the total number of trials (M&M's selected).
• $k$ is the number of successes (number of green M&M's).
• $p$ is the probability of success on a single trial (probability of selecting a green M&M).
• $\binom{n}{k}$ is the binomial coefficient, which is computed as $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

From the problem, we know that:

• $p = 0.15$ (probability of selecting a green M&M).
• $n = 5$ (since we select five M&M's).
• $k$ is the number of green M&M's, which varies in each part of the problem.

Part 1: Probability that exactly 4 of the 5 M&M’s are green

Here, $k = 4$, and the binomial probability formula becomes:

$P(X = 4) = \binom{5}{4} (0.15)^4 (1 - 0.15)^{5 - 4}$

First, calculate the binomial coefficient:

$\binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5}{1} = 5$

Now calculate the probability:

$P(X = 4) = 5 \times (0.15)^4 \times (0.85)^1$

$P(X = 4) = 5 \times 0.00050625 \times 0.85$

$P(X = 4) = 5 \times 0.0004303125$

$P(X = 4) = 0.0021516$

So, the probability that exactly four of the five M&M's are green is 0.0022 (rounded to four decimal places).

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Part 2: Probability that three or four of the five M&M’s are green

For this, we need to calculate the probability of $X = 3$ and $X = 4$ and then add them together.

Probability of exactly 3 green M&M's ($X = 3$):

$P(X = 3) = \binom{5}{3} (0.15)^3 (1 - 0.15)^{5 - 3}$

First, calculate the binomial coefficient:

$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$

Now calculate the probability:

$P(X = 3) = 10 \times (0.15)^3 \times (0.85)^2$

$P(X = 3) = 10 \times 0.003375 \times 0.7225$

$P(X = 3) = 10 \times 0.0024384375$

$P(X = 3) = 0.0243844$

So, $P(X = 3) = 0.0244$ (rounded to four decimal places).

Total Probability of three or four green M&M's:

$P(X = 3 \text{ or } 4) = P(X = 3) + P(X = 4)$

$P(X = 3 \text{ or } 4) = 0.0244 + 0.0022$

$P(X = 3 \text{ or } 4) = 0.0266$

Thus, the probability that three or four of the five M&M's are green is 0.0266.

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Would you like more details or have any further questions? Here are five related questions that might interest you:

1. How do binomial probabilities change with different sample sizes?
2. What is the probability of getting no green M&M's in the same scenario?
3. How would we compute the probability if we selected 10 M&M's instead of 5?
4. What if the probability of a green M&M was different, such as 20%?
5. Can we use the normal approximation to the binomial distribution in this case?

Tip: Always check the conditions (like sample size and probability) to see if you can approximate a binomial distribution with a normal distribution.