Answer :
To find the probability of rolling a 5 exactly three times in ten rolls of a six-sided number cube (i.e., a die), we need to use the binomial distribution formula. Here is a detailed, step-by-step solution to this problem:
1. Identify the Parameters:
- Total number of rolls, [tex]\( n = 10 \)[/tex]
- Number of successful outcomes (rolling a 5), [tex]\( k = 3 \)[/tex]
- Probability of a successful outcome on a single trial (rolling a 5), [tex]\( p = \frac{1}{6} \)[/tex]
- Probability of a failure on a single trial (not rolling a 5), [tex]\( q = 1 - p = \frac{5}{6} \)[/tex]
2. Calculate the Binomial Coefficient:
The binomial coefficient, represented as [tex]\( \binom{n}{k} \)[/tex], gives the number of ways to choose [tex]\( k \)[/tex] successes out of [tex]\( n \)[/tex] trials and is calculated using the formula:
[tex]\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \][/tex]
For [tex]\( n = 10 \)[/tex] and [tex]\( k = 3 \)[/tex]:
[tex]\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120 \][/tex]
3. Calculate the Probability Using the Binomial Distribution Formula:
The binomial distribution formula is given by:
[tex]\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \][/tex]
Substituting the identified values:
[tex]\[ P(X = 3) = \binom{10}{3} \cdot \left(\frac{1}{6}\right)^3 \cdot \left(\frac{5}{6}\right)^{10-3} \][/tex]
We can break it down further:
[tex]\[ P(X = 3) = 120 \cdot \left(\frac{1}{6}\right)^3 \cdot \left(\frac{5}{6}\right)^7 \][/tex]
4. Simplify the Expression:
Simplifying the probabilities of success and failure:
[tex]\[ \left(\frac{1}{6}\right)^3 = \frac{1}{216} \][/tex]
[tex]\[ \left(\frac{5}{6}\right)^7 \][/tex]
Plug these values into the probability formula:
[tex]\[ P(X = 3) = 120 \cdot \frac{1}{216} \cdot \left(\frac{5}{6}\right)^7 \][/tex]
5. Combine the Coefficient with the Probabilities:
Now we need to calculate the exact value of the resulting probability.
The resulting probability of rolling a 5 exactly three times in ten rolls of a six-sided die is approximately [tex]\( 0.15504535957425192 \)[/tex].
Thus, the correct choice and resulting value are from the third option:
[tex]\[ 10^c 3\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^4 \][/tex]
which corresponds to a binomial coefficient of 120 and a probability of [tex]\( 0.15504535957425192 \)[/tex].
1. Identify the Parameters:
- Total number of rolls, [tex]\( n = 10 \)[/tex]
- Number of successful outcomes (rolling a 5), [tex]\( k = 3 \)[/tex]
- Probability of a successful outcome on a single trial (rolling a 5), [tex]\( p = \frac{1}{6} \)[/tex]
- Probability of a failure on a single trial (not rolling a 5), [tex]\( q = 1 - p = \frac{5}{6} \)[/tex]
2. Calculate the Binomial Coefficient:
The binomial coefficient, represented as [tex]\( \binom{n}{k} \)[/tex], gives the number of ways to choose [tex]\( k \)[/tex] successes out of [tex]\( n \)[/tex] trials and is calculated using the formula:
[tex]\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \][/tex]
For [tex]\( n = 10 \)[/tex] and [tex]\( k = 3 \)[/tex]:
[tex]\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120 \][/tex]
3. Calculate the Probability Using the Binomial Distribution Formula:
The binomial distribution formula is given by:
[tex]\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \][/tex]
Substituting the identified values:
[tex]\[ P(X = 3) = \binom{10}{3} \cdot \left(\frac{1}{6}\right)^3 \cdot \left(\frac{5}{6}\right)^{10-3} \][/tex]
We can break it down further:
[tex]\[ P(X = 3) = 120 \cdot \left(\frac{1}{6}\right)^3 \cdot \left(\frac{5}{6}\right)^7 \][/tex]
4. Simplify the Expression:
Simplifying the probabilities of success and failure:
[tex]\[ \left(\frac{1}{6}\right)^3 = \frac{1}{216} \][/tex]
[tex]\[ \left(\frac{5}{6}\right)^7 \][/tex]
Plug these values into the probability formula:
[tex]\[ P(X = 3) = 120 \cdot \frac{1}{216} \cdot \left(\frac{5}{6}\right)^7 \][/tex]
5. Combine the Coefficient with the Probabilities:
Now we need to calculate the exact value of the resulting probability.
The resulting probability of rolling a 5 exactly three times in ten rolls of a six-sided die is approximately [tex]\( 0.15504535957425192 \)[/tex].
Thus, the correct choice and resulting value are from the third option:
[tex]\[ 10^c 3\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^4 \][/tex]
which corresponds to a binomial coefficient of 120 and a probability of [tex]\( 0.15504535957425192 \)[/tex].
