Answer :
To determine the probability of rolling a 4 exactly 2 times out of 7 rolls of a number cube, you can utilize the binomial probability formula:
[tex]\[ P(k \text{ successes }) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
Here's a step-by-step breakdown:
1. Identify the variables:
- [tex]\( n = 7 \)[/tex]: The total number of trials (rolls).
- [tex]\( k = 2 \)[/tex]: The number of successful trials (rolling a 4).
- [tex]\( p = \frac{1}{6} \)[/tex]: The probability of a single success since there is one 4 in six possible outcomes on a number cube.
- [tex]\( 1 - p = \frac{5}{6} \)[/tex]: The probability of a single failure (not rolling a 4).
2. Binomial coefficient:
The binomial coefficient [tex]\(\binom{n}{k}\)[/tex] represents the number of ways to choose [tex]\( k \)[/tex] successes out of [tex]\( n \)[/tex] trials. This is calculated as:
[tex]\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \][/tex]
For our specific case:
[tex]\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = 21 \][/tex]
3. Plug the values into the binomial formula:
[tex]\[ P(2 \text{ successes }) = \binom{7}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^5 \][/tex]
Substituting in the binomial coefficient and the probabilities:
[tex]\[ P(2 \text{ successes }) = 21 \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^5 \][/tex]
After calculating, you get:
- Binomial coefficient, [tex]\(\binom{7}{2}\)[/tex], is 21.
- The exact probability value (not calculated here per instruction) is approximately 0.2344285836762689.
Thus, the expression representing the probability of rolling a 4 exactly 2 times out of 7 rolls is:
[tex]\[ { }_7 C_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^5 \][/tex]
[tex]\[ P(k \text{ successes }) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
Here's a step-by-step breakdown:
1. Identify the variables:
- [tex]\( n = 7 \)[/tex]: The total number of trials (rolls).
- [tex]\( k = 2 \)[/tex]: The number of successful trials (rolling a 4).
- [tex]\( p = \frac{1}{6} \)[/tex]: The probability of a single success since there is one 4 in six possible outcomes on a number cube.
- [tex]\( 1 - p = \frac{5}{6} \)[/tex]: The probability of a single failure (not rolling a 4).
2. Binomial coefficient:
The binomial coefficient [tex]\(\binom{n}{k}\)[/tex] represents the number of ways to choose [tex]\( k \)[/tex] successes out of [tex]\( n \)[/tex] trials. This is calculated as:
[tex]\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \][/tex]
For our specific case:
[tex]\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = 21 \][/tex]
3. Plug the values into the binomial formula:
[tex]\[ P(2 \text{ successes }) = \binom{7}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^5 \][/tex]
Substituting in the binomial coefficient and the probabilities:
[tex]\[ P(2 \text{ successes }) = 21 \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^5 \][/tex]
After calculating, you get:
- Binomial coefficient, [tex]\(\binom{7}{2}\)[/tex], is 21.
- The exact probability value (not calculated here per instruction) is approximately 0.2344285836762689.
Thus, the expression representing the probability of rolling a 4 exactly 2 times out of 7 rolls is:
[tex]\[ { }_7 C_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^5 \][/tex]
