We are dealing with a binomial probability distribution because we are finding the probability of having exactly $r$ successes out of $n$ trials, where each trial has a success probability $p$ and a failure probability $q$.

The formula for the binomial probability is:

$P(r) = \binom{n}{r} \cdot p^r \cdot q^{n-r}$

Where:

• $\binom{n}{r}$ is the binomial coefficient, or "n choose r", which is calculated as $\frac{n!}{r!(n-r)!}$,
• $p$ is the probability of success,
• $q$ is the probability of failure ($q = 1 - p$),
• $n$ is the number of trials,
• $r$ is the number of successes.

Given values:

• $p = 0.18$,
• $q = 0.82$,
• $n = 4$,
• $r = 4$.

First, calculate $\binom{4}{4}$, which simplifies to 1 because $\binom{n}{n} = 1$.

Now, calculate the probability:

$P(4) = 1 \cdot (0.18)^4 \cdot (0.82)^0$

Let's calculate this result.The probability of exactly 4 successes in 4 trials, with a success probability of 0.18, is approximately 0.001 (rounded to the nearest thousandth).

Do you want more details or have any questions?

Here are some related questions:

1. How would the probability change if $r = 3$?
2. What is the probability of getting fewer than 4 successes?
3. How does increasing $p$ to 0.25 affect the result for $r = 4$?
4. What is the total probability of getting 0, 1, 2, or 3 successes?
5. How would you compute the expected number of successes?

Tip: The sum of probabilities over all possible values of