Question

if you roll a fair 6-sided die 9 times, what is the probability that at least 2 of the rolls come up as a 3 or a 4?

Answer 1

Using the binomial distribution, it is found that there is a 0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For each die, there are only two possible outcomes, either a 3 or a 4 is rolled, or it is not. The result of a roll is independent of any other roll, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• There are 9 rolls, hence $n = 9$.

• Of the six sides, 2 are 3 or 4, hence $p = \frac{2}{6} = 0.3333$

The desired probability is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1)$

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.3333)^{0}.(0.6667)^{9} = 0.026$

$P(X = 1) = C_{9,1}.(0.3333)^{1}.(0.6667)^{8} = 0.117$

Then:

$P(X < 2) = P(X = 0) + P(X = 1) = 0.026 + 0.117 = 0.143$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.143 = 0.857$

0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For more on the binomial distribution, you can check brainly.com/question/24863377