We roll a die 12 times. a) What is the probability that we get exactly 3 times the result greater than or equal to 4? .....

Question

We roll a die 12 times.

a) What is the probability that we get exactly 3 times the result greater than or equal to 4?

b) What is the probability of getting at most 3 times the result greater than or equal to 4?

The numbers greater than 4 are 4, 5, and 6, so the probability that in 1 toss we get a number greater than or equal to 4. Let's call this scenario a.

The numbers less than 4 are 1, 2 and 3, so the probability that in 1 toss we have a number less than 4. Let's call this scenario b.

When rolling 12 dice we have the sample space of $\frac{1}{6^{12}}$

When rolling the 12 dice, we have the probability of not getting any success (none of the rolls is greater than four), in this case the probability

$P[X = 0] = P[b,b,b,b,b,b,b,b,b,b,b,b ] = \binom{12}{0}=\frac{1}{6 ^{12}}$

The probability of getting 1 success is

$P[X = 1] = P[(a,b,b,b,b,b,b,b,b,b,b,b), (b,a,b,b,b,b,b,b,b,b,b,b),(b,b,a,b,b,b,b,b,b,b,b,b),(b,b,b,a,b,b,b,b,b,b,b,b),(b,b,b,b,a,b,b,b,b,b,b,b),(b,b,b,b,b, a,b,b,b,b,b,b),(b,b,b,b,b,b,a,b,b,b,b,b),(b,b,b,b,b,b,b,a,b,b,b,b),(b,b,b,b,b,b,b,b,a,b,b,b),(b,b,b,b,b,b,b,b,b,a,b,b),(b,b,b,b,b,b,b,b,b,b,a,b),(b,b,b) ,b,b,b,b,b,b,b,b,a)] = \frac{\binom{12}{1}}{6^{12}}= \frac{12 } {6^{ 12}}$

$P[X = 2] = \frac{\binom{12}{2}}{6^{12}} = \frac{66}{6^{12}}$

$P[X = 3] = \frac{\binom{12}{3}}{6^{12}} = \frac{220}{6^{12}}$

a-) $\frac{220}{6^{12}}$

b-) $\frac{1}{6^{12}} + \frac{12}{6^{12}} + \frac{66}{6^{12}} + \frac{220}{6^ {12}} = \frac{299}{6^{12}}$

I tried using random variables to solve the problem. Not sure if I was supposed to use random variables or if I used it correctly. But, am I on the right track?

Thanks for any help.

Answer 1

The binomial distribution states that if there are $n$ trials, each with probability $p$ of success, then the probability of exactly $k$ successes is $$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$ where $p^k$ is the probability of $k$ successes, $(1 - p)^{n - k}$ is the probability of $n - k$ failures, and $\binom{n}{k}$ is the number of ways exactly $k$ successes could occur in $n$ trials.

We roll a die $12$ times. What is the probability that we get exactly $3$ times a result greater than or equal to $4$?

Since there are $6$ possible results for each of the $12$ rolls of the die, there are $6^{12}$ elements in our sample space.

There are three possible successful results: $4, 5,$ or $6$. Assuming the die is fair, this means the probability of success is $$p = \frac{3}{6} = \frac{1}{2}$$

Thus, the probability of exactly three successes in $12$ trials is \begin{align*} \Pr(X = 3) & = \binom{12}{3}\left(\frac{1}{2}\right)^3\left(1 - \frac{1}{2}\right)^{12 - 3}\\ & = \binom{12}{3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^9\\ & = \binom{12}{3}\left(\frac{1}{2}\right)^{12} \end{align*} In your attempt, you did not take into account that there are three possible favorable results on each roll.

We roll a die $12$ times. What is the probability that we get at most $3$ times a result greater than or equal to $4$?

Again, assuming the die is fair, $p = 1/2$, the probability that at most $3$ rolls result in a number greater than or equal to $4$ is \begin{align*} \Pr(X \leq 3) & = \Pr(X = 0) + \Pr(X = 1) + \Pr(X = 2) + \Pr(X = 3)\\ & = \sum_{k = 0}^{3} \binom{12}{k}\left(\frac{1}{2}\right)^k\left(1 - \frac{1}{2}\right)^{12 - k}\\ & = \sum_{k = 0}^{3} \binom{12}{k}\left(\frac{1}{2}\right)^k\left(\frac{1}{2}\right)^{12 - k}\\ & = \sum_{k = 0}^{3} \binom{12}{k}\left(\frac{1}{2}\right)^{12}\\ & = \left(\frac{1}{2}\right)^{12}\sum_{k = 0}^{3} \binom{12}{k} \end{align*} I will leave the remainder of the calculation to you.