Using the binomial distribution, we have that:
a) There is a 0.2111 = 21.11% probability that a six will occur exactly twice.
b) The expected number of sixes is of 3.2.
c) The variance of the number of sixes is of 2.35.
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 parameters are given by:
p = 8/30 = 0.2667, n = 12.
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_{12,2}.(0.2667)^{2}.(1-0.2667)^{10} = 0.2111[/tex]
Item b:
The expected number of the binomial distribution is:
E(X) = np.
Hence:
E(X) = 12 x 8/30 = 3.2.
Item c:
The variance of the binomial distribution is:
V(X) = np(1-p).
Hence:
E(X) = 12 x 8/30 x 22/30 = 2.35.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
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