Answer:
a) 0.2182
b) 0.0691
c) 0.9309
Step-by-step explanation:
Binomial probability distribution
The binomial probability is the probability of exactly b successes on n repeated trials, and B can only have two outcomes.
[tex]P(B = b) = C_{n,b}.p^{b}.(1-p)^{n-b}[/tex]
In which [tex]C_{n,b}[/tex] is the number of different combinations of b objects from a set of n elements, given by the following formula.
[tex]C_{n,b} = \frac{n!}{x!(n-b)!}[/tex]
And p is the probability of B happening.
In this problem we have that:
Bin(20,0.2).
This means that [tex]n = 20, p = 0.2[/tex]
(a) P(B=4).
[tex]P(B = b) = C_{n,b}.p^{b}.(1-p)^{n-b}[/tex]
[tex]P(B = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182[/tex]
(b) P(B≤1).
[tex]P(B \leq 1) = P(B = 0) + P(B = 1)[/tex]
[tex]P(B = b) = C_{n,b}.p^{b}.(1-p)^{n-b}[/tex]
[tex]P(B = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115[/tex]
[tex]P(B = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576[/tex]
[tex]P(B \leq 1) = P(B = 0) + P(B = 1) = 0.0115 + 0.0576 = 0.0691[/tex]
(c) P(B>1).
Either B is less than or equal to 1, or B is larger than 1. The sum of the probabilities of these events is decimal 1. So
[tex]P(B \leq 1) + P(B > 1) = 1[/tex]
We have that, from b), [tex]P(B \leq 1) = 0.0691[/tex]
So
[tex]0.0691 + P(B > 1) = 1[/tex]
[tex]P(B > 1) = 0.9309[/tex]
