To solve this problem, we will use the binomial probability formula:

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

where:

• $n = 5$ is the number of trials,
• $p = 0.628$ is the probability of success on each trial,
• $k$ is the number of successes, which ranges from 0 to 5,
• $\binom{n}{k}$ is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$.

We'll calculate $P(X = k)$ for $k = 0, 1, 2, 3, 4,$ and $5$, and round each answer to three decimal places.

Calculations:

1. For $k = 0$: $P(X = 0) = \binom{5}{0} (0.628)^0 (1 - 0.628)^5$

2. For $k = 1$: $P(X = 1) = \binom{5}{1} (0.628)^1 (1 - 0.628)^4$

3. For $k = 2$: $P(X = 2) = \binom{5}{2} (0.628)^2 (1 - 0.628)^3$

4. For $k = 3$: $P(X = 3) = \binom{5}{3} (0.628)^3 (1 - 0.628)^2$

5. For $k = 4$: $P(X = 4) = \binom{5}{4} (0.628)^4 (1 - 0.628)^1$

6. For $k = 5$: $P(X = 5) = \binom{5}{5} (0.628)^5 (1 - 0.628)^0$

Let’s calculate each probability.The probabilities for each value of $k$ are as follows:

• $P(X = 0) = 0.007$
• $P(X = 1) = 0.060$
• $P(X = 2) = 0.203$
• $P(X = 3) = 0.343$
• $P(X = 4) = 0.289$
• $P(X = 5) = 0.098$

Let me know if you would like further details on any of the calculations or have other questions.

Related Questions:

1. How is the binomial coefficient $\binom{n}{k}$ calculated?
2. How does changing $p$ affect the binomial distribution?
3. What are the properties of a binomial distribution?
4. How does the binomial distribution approach a normal distribution for large $n$?
5. How would the probabilities change if $n$ were different?

Tip:

In a binomial distribution, if $p$ is close to 0.5 and $n$ is large, the distribution is approximately symmetric and resembles a normal distribution.