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

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

where:

• $n = 4$: the number of trials,
• $p = 0.25$: the probability of success on each trial,
• $k = 2$: the number of successes we're interested in.

Step-by-Step Solution:

1. Calculate $\binom{n}{k}$: $\binom{n}{k} = \frac{n!}{k!(n - k)!} = \frac{4!}{2!(4 - 2)!} = \frac{4 \times 3}{2 \times 1} = 6$

2. Apply the binomial probability formula: Substitute the values into the formula: $P(X = 2) = 6 \times (0.25)^2 \times (1 - 0.25)^{4 - 2}$

3. Compute the probability:

   • $(0.25)^2 = 0.0625$
   • $(1 - 0.25) = 0.75$
   • $0.75^{2} = 0.5625$

   So, $P(X = 2) = 6 \times 0.0625 \times 0.5625 = 0.2109375$

4. Round the answer to three decimal places: $P(X = 2) \approx 0.211$

Answer:

The probability that the experiment results in exactly 2 successes is 0.211.

Would you like more details or have any questions?

Related Questions:

1. How would the probability change if the number of trials $n$ was increased to 6 with the same probability of success?
2. What is the probability of getting exactly 1 success in this scenario?
3. How does the probability change if the success probability $p$ is increased to 0.5?
4. What is the probability of having 0 successes with $n = 4$ and $p = 0.25$?
5. How can we calculate the cumulative probability of getting at most 2 successes?

Tip:

When solving binomial problems, double-check the values of $n$, $k$, and $p$ to ensure they’re substituted correctly, as small errors in setup can lead to incorrect answers.