Question

find the probability that you will roll an even number exactly 5 times when you: roll a six - sided number cube 10 times. $p = square$ roll a six - sided number cube 20 times. $p = square$

Explanation:

Step1: Identify the binomial - probability formula

The binomial - probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successful trials, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. For rolling a six - sided die, the probability of rolling an even number ($2$, $4$, or $6$) is $p=\frac{3}{6}=\frac{1}{2}$, and $1 - p=\frac{1}{2}$.

Step2: Calculate for $n = 10$ and $k = 5$

First, calculate the combination $C(10,5)=\frac{10!}{5!(10 - 5)!}=\frac{10!}{5!5!}=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1}=252$. Then, $p=\frac{1}{2}$, $k = 5$, $n = 10$, and $1 - p=\frac{1}{2}$. So $P(X = 5)=C(10,5)\times(\frac{1}{2})^{5}\times(\frac{1}{2})^{10 - 5}=252\times(\frac{1}{2})^{10}=252\times\frac{1}{1024}=\frac{252}{1024}=\frac{63}{256}\approx0.246$.

Step3: Calculate for $n = 20$ and $k = 5$

Calculate the combination $C(20,5)=\frac{20!}{5!(20 - 5)!}=\frac{20\times19\times18\times17\times16}{5\times4\times3\times2\times1}=15504$. Then, $p=\frac{1}{2}$, $k = 5$, $n = 20$, and $1 - p=\frac{1}{2}$. So $P(X = 5)=C(20,5)\times(\frac{1}{2})^{5}\times(\frac{1}{2})^{20 - 5}=15504\times(\frac{1}{2})^{20}=15504\times\frac{1}{1048576}=\frac{15504}{1048576}=\frac{969}{65536}\approx0.0148$.

Answer:

For rolling a six - sided number cube 10 times: $\frac{63}{256}$
For rolling a six - sided number cube 20 times: $\frac{969}{65536}$