Using the binomial distribution, the probabilities are given as follows:
a) 0.4159 = 41.59%.
b) 0.5610 = 56.10%.
c) 0.8549 = 85.49%.
The formula is:
[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:
For this problem, the values of the parameters are:
n = 3, p = 0.76.
Item a:
The probability is P(X = 2), hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{3,2}.(0.76)^{2}.(0.24)^{1} = 0.4159[/tex]
Item b:
The probability is P(X < 3), hence:
P(X < 3) = 1 - P(X = 3)
In which:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{3,3}.(0.76)^{3}.(0.24)^{0} = 0.4390[/tex]
Then:
P(X < 3) = 1 - P(X = 3) = 1 - 0.4390 = 0.5610 = 56.10%.
Item c:
The probability is:
[tex]P(X \geq 2) = P(X = 2) + P(X = 3) = 0.4159 + 0.4390 = 0.8549[/tex]
More can be learned about the binomial distribution at https://brainly.com/question/24863377
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