Respuesta :
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
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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 [tex]n = 9[/tex].
- Of the six sides, 2 are 3 or 4, hence [tex]p = \frac{2}{6} = 0.3333[/tex]
The desired probability is:
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which:
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
Then
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{9,0}.(0.3333)^{0}.(0.6667)^{9} = 0.026[/tex]
[tex]P(X = 1) = C_{9,1}.(0.3333)^{1}.(0.6667)^{8} = 0.117[/tex]
Then:
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.026 + 0.117 = 0.143[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.143 = 0.857[/tex]
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 https://brainly.com/question/24863377
