Thuy rolls a number cube 7 times. Which expression represents the probability of rolling a 4 exactly 2 times?

A. [tex]\({ }_7 C_5\left(\frac{1}{6}\right)^2\left(\frac{1}{8}\right)^5\)[/tex]
B. [tex]\({ }_7 C_5\left(\frac{1}{6}\right)^5\left(\frac{5}{6}\right)^2\)[/tex]
C. [tex]\({ }_7 C_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^5\)[/tex]



Answer :

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]