Answered

At IDNLearn.com, find answers to your most pressing questions from experts and enthusiasts alike. Our platform provides accurate, detailed responses to help you navigate any topic with ease.

If a fair die is rolled 7 times, what is the probability, rounded to the nearest thousandth, of getting at most 2 twos?

Sagot :

Avan8idn

Answer:

0.904

Step-by-step explanation:

Using the binomial distribution, it is found that there is a 0.905 = 90.5% probability of getting at most 2 twos.

For each time the die is rolled, there are only two possible outcomes, either the result is a 2, or it is not. The result on a roll is independent of any other roll, which means that 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:

  • The die is rolled 7 times, hence [tex]n = 7[/tex].
  • One out of the 6 sides is a 2, hence [tex]p = \frac{1}{6} = 0.1667[/tex]

The probability is:

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

Hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{7,0}.(0.1667)^{0}.(0.8333)^{7} = 0.279[/tex]

[tex]P(X = 1) = C_{7,1}.(0.1667)^{1}.(0.8333)^{6} = 0.391[/tex]

[tex]P(X = 2) = C_{7,2}.(0.1667)^{2}.(0.8333)^{5} = 0.235[/tex]

Then:

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.279 + 0.391 + 0.235 = 0.905[/tex]

0.905 = 90.5% probability of getting at most 2 twos.

A similar problem is given at https://brainly.com/question/24863377

We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Your questions deserve precise answers. Thank you for visiting IDNLearn.com, and see you again soon for more helpful information.