To determine the probability of rolling a 4 exactly 2 times out of 7 rolls of a number cube (where each side is equally likely), we will use the binomial probability formula:
[tex]\[
P(k \text{ successes}) = \binom{n}{k} p^k (1-p)^{n-k}
\][/tex]
Here, [tex]\( n = 7 \)[/tex], [tex]\( k = 2 \)[/tex], [tex]\( p = \frac{1}{6} \)[/tex] (probability of rolling a 4), and [tex]\( q = 1 - p = \frac{5}{6} \)[/tex] (probability of not rolling a 4).
1. Binomial Coefficient Calculation:
[tex]\[
\binom{n}{k} = \binom{7}{2} = \frac{7!}{(7-2)!2!} = \frac{7!}{5!2!} = \frac{7 \times 6}{2 \times 1} = 21
\][/tex]
2. Probability Calculation:
[tex]\[
P(2 \text{ successes}) = 21 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{5}
\][/tex]
Given the set of expressions:
1. [tex]\(\binom{7}{5}\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^5\)[/tex]
2. [tex]\(\binom{7}{5}\left(\frac{1}{6}\right)^5\left(\frac{5}{6}\right)^2\)[/tex]
3. [tex]\(\binom{7}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^5\)[/tex]
From the calculations:
- The correct binomial coefficient should be [tex]\(\binom{7}{2}\)[/tex], not [tex]\(\binom{7}{5}\)[/tex].
- The powers of [tex]\(\frac{1}{6}\)[/tex] and [tex]\(\frac{5}{6}\)[/tex] should match the formulas used above.
Thus, the correct expression from the given choices is:
[tex]\[
\boxed{\binom{7}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^5}
\][/tex]
