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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
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