The problem involves calculating the probability of rolling a 4 exactly 2 times when a number cube (die) is rolled 7 times. We'll follow these steps:
1. Identify the components of the binomial probability formula:
[tex]\[
P(k \text{ successes}) = { }_n C_k \, p^k \, (1-p)^{n-k}
\][/tex]
where:
- [tex]\( n \)[/tex] is the number of trials (7 rolls of the die).
- [tex]\( k \)[/tex] is the number of successes (rolling a 4 exactly 2 times).
- [tex]\( p \)[/tex] is the probability of success on a single trial (rolling a 4), which is [tex]\(\frac{1}{6}\)[/tex].
- [tex]\( 1 - p \)[/tex] is the probability of failure on a single trial (not rolling a 4), which is [tex]\(\frac{5}{6}\)[/tex].
2. Identify the binomial coefficient [tex]\( { }_n C_k \)[/tex]:
[tex]\[
{ }_n C_k = \frac{n!}{k!(n-k)!}
\][/tex]
For this problem, [tex]\( { }_7 C_2 \)[/tex]:
[tex]\[
{ }_7 C_2 = \frac{7!}{2!(7-2)!} = \frac{7!}{2! \cdot 5!}
\][/tex]
3. Construct the binomial probability expression:
[tex]\[
P(k=2) = { }_7 C_2 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^5
\][/tex]
4. Plug in the values and identify the correct choice:
[tex]\[
{ }_7 C_2 \, \left(\frac{1}{6}\right)^2 \, \left(\frac{5}{6}\right)^5
\][/tex]
The correct expression representing the probability of rolling a 4 exactly 2 times in 7 rolls is:
[tex]\[
{ }_7 C_2 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^5
\][/tex]
