Question

The probability of rolling a 4 on a six-sided number cube is
6
1
​
. What is the probability of rolling a 4 twice out of five total rolls?
Use the formula
n
​
C
k
​
p
k
q
n−k
to solve. Write your answer as a whole number probability.
□ percent

Answer 1

To find the probability of rolling a 4 twice in five rolls of a six-sided die, we can use the provided formula:

\[
P(X = k) = nCk \cdot p^k \cdot q^{n-k}
\]

Where:
- \( n \) = total number of trials (rolls) = 5
- \( k \) = number of successes (rolling a 4) = 2
- \( p \) = probability of success on a single trial = \(\frac{1}{6}\) (probability of rolling a 4)
- \( q \) = probability of failure (not rolling a 4) = \(1 - p = \frac{5}{6}\)

First, we need to calculate \( nCk \):

\[
nCk = \frac{n!}{k!(n-k)!} = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \cdot 3!} = \frac{5 \cdot 4}{2 \cdot 1} = 10
\]

Now we can substitute into the formula:

\[
P(X = 2) = nCk \cdot p^k \cdot q^{n-k}
\]
\[
= 10 \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{3}
\]

Calculating \( \left(\frac{1}{6}\right)^2 \):

\[
\left(\frac{1}{6}\right)^2 = \frac{1}{36}
\]

Calculating \( \left(\frac{5}{6}\right)^3 \):

\[
\left(\frac{5}{6}\right)^3 = \frac{125}{216}
\]

Now substituting back into the probability equation:

\[
P(X = 2) = 10 \cdot \frac{1}{36} \cdot \frac{125}{216}
\]

Calculating the product:

\[
= 10 \cdot \frac{125}{7776} = \frac{1250}{7776}
\]

Now, to convert this to a percentage, first calculate it as a decimal:

\[
\frac{1250}{7776} \approx 0.16077
\]

To convert this to a percentage, multiply by 100:

\[
0.16077 \times 100 \approx 16.08
\]

Rounded to the nearest whole number, the probability of rolling a 4 twice out of five total rolls is approximately:

\[
\text{□ percent} \approx 16\%
\]