Respuesta :
Answer:
a) P(X = 0) = 0.5997
b) P(X = 9) = 0.0016
c) P(X = 8) = 0.0047
d) P(X = 5) = 0.4018
Step-by-step explanation:
These following problem are examples of the binomial probability distribution.
Binomial probability
Th binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And [tex]\pi[/tex] is the probability of X happening.
(a) for n = 4 and π = 0.12, what is P(X = 0)?
[tex]P(X = 0) = C_{4,0}.(0.12)^{0}.(0.88)^{4} = 0.5997[/tex]
(b) for n = 10 and π = 0.40, what is P(X = 9)?
[tex]P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016[/tex]
(c) for n = 10 and π = 0.50, what is P(X = 8)?
[tex]P(X = 8) = C_{10,8}.(0.5)^{8}.(0.5)^{2} = 0.0047[/tex]
(d) for n = 6 and π = 0.83, what is P(X = 5)?
[tex]P(X = 5) = C_{6,5}.(0.83)^{5}.(0.17)^{1} = 0.4018[/tex]
The required answers are,
Part(a):[tex]P(X=0)=0.600[/tex]
Part(b):[tex]P(X=9)=0.0.002[/tex]
Part(c): [tex]P(X=8)=0.044[/tex]
Part(d): [tex]P(X=5)=0.402[/tex]
Binomial distribution:
The formula for the binomial distribution is,
[tex]P(x)=n_C_x p^x(1-p)^{n-x}[/tex]
Part(a):
Given that,
[tex]n=4,\pi=0.12[/tex]
Now, substituting the given values into the above formula we get,
[tex]P(X=0)=\left[\begin{array}{ccc}n\\x\\\end{array}\right] \pi^x(1-\pi)^{n-x}\\=\left[\begin{array}{ccc}4\\0\\\end{array}\right] (0.12)^0(1-0.12)^{4-0}\\=0.600[/tex]
Part(b):
Given that,
[tex]n=9,\pi=0.4[/tex]
Now, substituting the given values into the above formula we get,
[tex]P(X=9)=\left[\begin{array}{ccc}n\\x\\\end{array}\right] \pi^x(1-\pi)^{n-x}\\=\left[\begin{array}{ccc}10\\9\\\end{array}\right] (0.4)^9(1-0.4)^{10-9}\\=0.002[/tex]
Part(c):
Given that,
[tex]n=8,\pi=0.5[/tex]
Now, substituting the given values into the above formula we get,
[tex]P(X=8)=\left[\begin{array}{ccc}n\\x\\\end{array}\right] \pi^x(1-\pi)^{n-x}\\=\left[\begin{array}{ccc}10\\8\\\end{array}\right] (0.5)^9(1-0.5)^{10-8}\\=0.044[/tex]
Part(d):
Given that,
[tex]n=6,\pi=0.83[/tex]
Now, substituting the given values into the above formula we get,
[tex]P(X=5)=\left[\begin{array}{ccc}n\\x\\\end{array}\right] \pi^x(1-\pi)^{n-x}\\=\left[\begin{array}{ccc}6\\5\\\end{array}\right] (0.83)^5(1-0.83)^{6-5}\\=0.402[/tex]
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