A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment.
n equals 9​, p equals 0.8​, x less than or equals 3
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Part 1
The probability of x less than or equals 3 successes is

enter your response here. ​(Round to four decimal places as​ needed.)

To calculate the probability of \( x \leq 3 \) successes in a binomial experiment, we use the binomial probability formula:

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

where:
- \( n = 9 \) (the number of trials),
- \( p = 0.8 \) (the probability of success on a single trial),
- \( 1 - p = 0.2 \) (the probability of failure),
- \( k \) is the number of successes.

The goal is to calculate the probability of having **3 or fewer** successes, meaning we need to compute the sum of probabilities for \( X = 0, X = 1, X = 2, \) and \( X = 3 \):

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

### Step-by-step calculation:

1. **For \( X = 0 \):**
   \[
   P(X = 0) = \binom{9}{0} (0.8)^0 (0.2)^9 = 1 \times 1 \times 0.000000512 = 0.000000512
   \]

2. **For \( X = 1 \):**
   \[
   P(X = 1) = \binom{9}{1} (0.8)^1 (0.2)^8 = 9 \times 0.8 \times 0.0000065536 = 0.0000471808
   \]

3. **For \( X = 2 \):**
   \[
   P(X = 2) = \binom{9}{2} (0.8)^2 (0.2)^7 = 36 \times 0.64 \times 0.00008192 = 0.0018905088
   \]

4. **For \( X = 3 \):**
   \[
   P(X = 3) = \binom{9}{3} (0.8)^3 (0.2)^6 = 84 \times 0.512 \times 0.000262144 = 0.01128192
   \]

### Adding all probabilities together:

\[
P(X \leq 3) = 0.000000512 + 0.0000471808 + 0.0018905088 + 0.01128192 = 0.0132201
\]

### Final Answer:
\[
P(X \leq 3) \approx 0.0132
\]

Thus, the probability of \( X \leq 3 \) successes is approximately **0.0132**.

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Do you have any questions, or would you like further details?

### Related Questions:
1. How does the binomial distribution differ from the normal distribution?
2. What is the probability of getting exactly 3 successes?
3. How can the cumulative distribution function (CDF) help in binomial probability calculations?
4. If \( p = 0.5 \), what would the probability \( P(X \leq 3) \) be in this case?
5. How does changing \( n \) to 10 affect the probability of \( X \leq 3 \)?

### Tip:
For binomial probability problems, if \( n \) is large, you can approximate the binomial distribution using the normal distribution for easier calculations.

A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment. n equals 9, p equals 0.8, x less than or equals 3. The probability of x less than or equals 3 successes is (Round to four decimal places as needed).

Compute the probability of x successes in a binomial probability experiment where n = 9 and p = 0.8, with x ≤ 3. Detailed step-by-step solution provided, including key formulas and theorems. Suitable for students in Grades 9-12.

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