Answer:
[tex]P(X = 0) = 0.0263[/tex]
[tex]P(X = 1) = 0.1407[/tex]
[tex]P(X = 2) = 0.3012[/tex]
[tex]P(X = 3) = 0.3224[/tex]
[tex]P(X = 4) = 0.1725[/tex]
[tex]P(X = 5) = 0.0369[/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 problem we have that:
[tex]n = 5, p = 0.517[/tex]
Distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{5,0}.(0.517)^{0}.(0.483)^{5} = 0.0263[/tex]
[tex]P(X = 1) = C_{5,1}.(0.517)^{1}.(0.483)^{4} = 0.1407[/tex]
[tex]P(X = 2) = C_{5,2}.(0.517)^{2}.(0.483)^{3} = 0.3012[/tex]
[tex]P(X = 3) = C_{5,3}.(0.517)^{3}.(0.483)^{2} = 0.3224[/tex]
[tex]P(X = 4) = C_{5,4}.(0.517)^{4}.(0.483)^{1} = 0.1725[/tex]
[tex]P(X = 5) = C_{5,5}.(0.517)^{5}.(0.483)^{0} = 0.0369[/tex]
