Certainly! Let's solve the problem step by step using the binomial probability formula.
Given:
- Number of trials (\( n \)) = 16
- Number of successes (\( x \)) = 2
- Probability of success on a given trial (\( p \)) = \( \frac{1}{4} \)
We use the binomial probability formula:
[tex]\[
P(X = x) = \binom{n}{x} \cdot p^x \cdot (1-p)^{n-x}
\][/tex]
Here, \( \binom{n}{x} \) is the binomial coefficient, which is calculated as:
[tex]\[
\binom{n}{x} = \frac{n!}{x!(n-x)!}
\][/tex]
Substitute the given values into the binomial coefficient formula:
[tex]\[
\binom{16}{2} = \frac{16!}{2!(16-2)!} = \frac{16!}{2! \cdot 14!}
\][/tex]
After canceling out common factorial terms, this simplifies to:
[tex]\[
\binom{16}{2} = \frac{16 \times 15}{2 \times 1} = 120
\][/tex]
Next, calculate the probability terms:
[tex]\[
p^x = \left(\frac{1}{4}\right)^2 = \frac{1}{16}
\][/tex]
[tex]\[
(1-p)^{n-x} = \left(1 - \frac{1}{4}\right)^{16-2} = \left(\frac{3}{4}\right)^{14}
\][/tex]
Now, putting it all together in the binomial probability formula:
[tex]\[
P(X = 2) = 120 \cdot \frac{1}{16} \cdot \left(\frac{3}{4}\right)^{14}
\][/tex]
After evaluating the arithmetic:
[tex]\[
P(X = 2) = 120 \cdot \frac{1}{16} \cdot 0.0953 \quad \text{(rounded to 4 decimal places for simplicity)}
\][/tex]
Finally, multiplying all factors together gives:
[tex]\[
P(X = 2) \approx 0.13363461010158062
\][/tex]
Rounded to three decimal places, the probability is:
[tex]\[
P(X = 2) \approx 0.134
\][/tex]
So, the probability of getting exactly 2 successes in 16 trials, with each trial having a success probability of \( \frac{1}{4} \), is approximately 0.134.
[tex]\[
\boxed{0.134}
\][/tex]
