Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
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
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
