Question

Assume that a procedure yields a binomial distribution with a trial repeated n=5
times. Use some form of technology to find the probability distribution given the probability p=0.636 of success on a single trial.

(Report answers accurate to 4 decimal places.)

k P(X = k)
0
1
2
3
4
5

Answer 1

To find the probability distribution of a binomial random variable with parameters \( n = 5 \) (number of trials) and \( p = 0.636 \) (probability of success on a single trial), we can use the binomial probability formula:

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

where:
- \( \binom{n}{k} \) is the binomial coefficient,
- \( p \) is the probability of success,
- \( 1-p \) is the probability of failure.

We will calculate \( P(X = k) \) for \( k = 0, 1, 2, 3, 4, 5 \).

First, calculate \( 1 - p = 1 - 0.636 = 0.364 \).

Now, we can compute each probability:

1. **For \( k = 0 \)**:
\[
P(X = 0) = \binom{5}{0} (0.636)^0 (0.364)^5 = 1 \cdot 1 \cdot (0.364)^5 \approx 0.0048
\]

2. **For \( k = 1 \)**:
\[
P(X = 1) = \binom{5}{1} (0.636)^1 (0.364)^4 = 5 \cdot (0.636) \cdot (0.364)^4 \approx 0.0588
\]

3. **For \( k = 2 \)**:
\[
P(X = 2) = \binom{5}{2} (0.636)^2 (0.364)^3 = 10 \cdot (0.636)^2 \cdot (0.364)^3 \approx 0.2246
\]

4. **For \( k = 3 \)**:
\[
P(X = 3) = \binom{5}{3} (0.636)^3 (0.364)^2 = 10 \cdot (0.636)^3 \cdot (0.364)^2 \approx 0.3844
\]

5. **For \( k = 4 \)**:
\[
P(X = 4) = \binom{5}{4} (0.636)^4 (0.364)^1 = 5 \cdot (0.636)^4 \cdot (0.364) \approx 0.2845
\]

6. **For \( k = 5 \)**:
\[
P(X = 5) = \binom{5}{5} (0.636)^5 (0.364)^0 = 1 \cdot (0.636)^5 \cdot 1 \approx 0.0858
\]

Finally, rounding the probabilities to four decimal places, we get:

\[
\begin{array}{c|c}
k & P(X = k) \\
\hline
0 & 0.0048 \\
1 & 0.0588 \\
2 & 0.2246 \\
3 & 0.3844 \\
4 & 0.2845 \\
5 & 0.0858 \\
\end{array}
\]

So the probability distribution is as follows:

- \( P(X = 0) \approx 0.0048 \)
- \( P(X = 1) \approx 0.0588 \)
- \( P(X = 2) \approx 0.2246 \)
- \( P(X = 3) \approx 0.3844 \)
- \( P(X = 4) \approx 0.2845 \)
- \( P(X = 5) \approx 0.0858 \)