Question

a die is rolled 7 times. find the probability of rolling the following.
exactly 7 threes.
use the formula for binomial probability to calculate the probability of rolling exactly 7 threes. assume success in this
case is rolling a three.
p(x successes in n trials) = c(n,x)·p^x·(1 - p)^{n - x}=c(7,7)·\frac{1}{6}^7·(1 - \frac{1}{6})^7
(type integers or fractions.)

Explanation:

Step1: Calculate combination \(C(n,x)\)

The combination formula \(C(n,x)=\frac{n!}{x!(n - x)!}\). Here \(n = 7\), \(x=7\).
\(C(7,7)=\frac{7!}{7!(7 - 7)!}=\frac{7!}{7!0!}=1\) (since \(0!=1\))

Step2: Calculate \(p^{x}\)

Given \(p=\frac{1}{6}\), \(x = 7\). So \(p^{x}=(\frac{1}{6})^{7}=\frac{1}{6^{7}}=\frac{1}{279936}\)

Step3: Calculate \((1 - p)^{n - x}\)

Given \(p=\frac{1}{6}\), \(n = 7\), \(x = 7\). Then \(1-p = 1-\frac{1}{6}=\frac{5}{6}\), \(n - x=7 - 7 = 0\). So \((1 - p)^{n - x}=(\frac{5}{6})^{0}=1\) (since \(a^{0}=1,a
eq0\))

Step4: Calculate the binomial probability

\(P(x = 7)=C(7,7)\times p^{7}\times(1 - p)^{0}\)
Substitute the values: \(P(x = 7)=1\times\frac{1}{279936}\times1=\frac{1}{279936}\)

Answer:

\(\frac{1}{279936}\)