Respuesta :
Answer:
[tex]P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}[/tex]
Possible values of x: Any from 0 to 5.
[tex]P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807[/tex]
[tex]P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015[/tex]
[tex]P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087[/tex]
[tex]P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323[/tex]
[tex]P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835[/tex]
[tex]P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243[/tex]
Step-by-step explanation:
Binomial probability distribution
The 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}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations 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 p is the probability of X happening.
In this question:
[tex]n = 5, p = 0.3, q = 1 - p = 0.7[/tex]
So
[tex]P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}[/tex]
Possible values of x: 5 trials, so any value from 0 to 5.
For each value of x calculate p(☓ =x)
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807[/tex]
[tex]P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015[/tex]
[tex]P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087[/tex]
[tex]P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323[/tex]
[tex]P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835[/tex]
[tex]P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243[/tex]
