find the probability of r successes and round to the nearest thousandth if p = 0.18, q = 0.82, n = 4, and r = 4

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

Learn how to calculate the probability of exactly 4 successes in 4 trials using binomial probability distribution with p = 0.18 and n = 4. Step-by-step explanation includes the binomial formula and rounding to the nearest thousandth. Suitable for students in Grades 10-12.

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