Constructed binomial distribution using N equals six and P equals .32

To construct a binomial distribution with \( N = 6 \) (number of trials) and \( P = 0.32 \) (probability of success for each trial), we can calculate the probability of obtaining \( k \) successes out of \( N \) trials using the binomial probability formula:

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

Where:
- \( P(X = k) \) is the probability of getting \( k \) successes,
- \( \binom{N}{k} = \frac{N!}{k!(N-k)!} \) is the binomial coefficient,
- \( N = 6 \),
- \( P = 0.32 \), and
- \( 1 - P = 0.68 \) (probability of failure).

We will compute the probabilities for \( k = 0, 1, 2, 3, 4, 5, 6 \).

### Step-by-step calculation:
1. **For \( k = 0 \):**
\[
P(X = 0) = \binom{6}{0} (0.32)^0 (0.68)^6 = 1 \times 1 \times 0.68^6 = 0.0886
\]

2. **For \( k = 1 \):**
\[
P(X = 1) = \binom{6}{1} (0.32)^1 (0.68)^5 = 6 \times 0.32 \times 0.68^5 = 0.2498
\]

3. **For \( k = 2 \):**
\[
P(X = 2) = \binom{6}{2} (0.32)^2 (0.68)^4 = 15 \times 0.32^2 \times 0.68^4 = 0.3313
\]

4. **For \( k = 3 \):**
\[
P(X = 3) = \binom{6}{3} (0.32)^3 (0.68)^3 = 20 \times 0.32^3 \times 0.68^3 = 0.2226
\]

5. **For \( k = 4 \):**
\[
P(X = 4) = \binom{6}{4} (0.32)^4 (0.68)^2 = 15 \times 0.32^4 \times 0.68^2 = 0.0886
\]

6. **For \( k = 5 \):**
\[
P(X = 5) = \binom{6}{5} (0.32)^5 (0.68)^1 = 6 \times 0.32^5 \times 0.68 = 0.0171
\]

7. **For \( k = 6 \):**
\[
P(X = 6) = \binom{6}{6} (0.32)^6 (0.68)^0 = 1 \times 0.32^6 \times 1 = 0.0011
\]

### Summary of the binomial distribution:
| \( k \) (Number of successes) | Probability \( P(X = k) \) |
|-------------------------------|----------------------------|
| 0                             | 0.0886                     |
| 1                             | 0.2498                     |
| 2                             | 0.3313                     |
| 3                             | 0.2226                     |
| 4                             | 0.0886                     |
| 5                             | 0.0171                     |
| 6                             | 0.0011                     |

This table provides the probabilities for each possible number of successes when performing 6 trials, with a success probability of 0.32.

Would you like more details or have any questions?

### Related questions:
1. How would the distribution change if \( P \) were 0.5 instead of 0.32?
2. What is the expected value (mean) and variance of this binomial distribution?
3. How can we plot the binomial distribution for \( N = 6 \) and \( P = 0.32 \)?
4. What is the probability of getting at most 2 successes?
5. How does the shape of the binomial distribution change as \( N \) increases?

### Tip:
In a binomial distribution, the mean is \( \mu = N \times P \) and the variance is \( \sigma^2 = N \times P \times (1 - P) \).

Learn how to construct a binomial distribution with 6 trials (N = 6) and a probability of success (P) of 0.32. This detailed guide explains how to calculate the probabilities of obtaining k successes and covers key concepts such as binomial probability formula and the binomial theorem. Suitable for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation