Answer:
P(X ≤ 3) = 0.9933.
Step-by-step explanation:
We are given that the random variable X has a binomial distribution with the given probability of obtaining a success
Also, given n = 5, p =0.2.
The above situation can be represented through binomial distribution;
[tex]P(X = r)= \binom{n}{r} \times p^{r} \times (1-p)^{n-r} ;x = 0,1,2,3,.........[/tex]
where, n = number of samples (trials) taken = 5
r = number of success = less than equal to 3
p = probability of success which in our question is 0.20.
Let X = A random variable
So, X ~ Binom(n = 5, p = 0.20)
Now, the probability that X is less than and equal to 3 is given by = P(X ≤ 3)
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= [tex]\binom{5}{0} \times 0.20^{0} \times (1-0.20)^{5-0}+\binom{5}{1} \times 0.20^{1} \times (1-0.20)^{5-1}+\binom{5}{2} \times 0.20^{2} \times (1-0.20)^{5-2}+\binom{5}{3} \times 0.20^{3} \times (1-0.20)^{5-3}[/tex]
= [tex]1 \times 1 \times 0.80^{5}+5 \times 0.20^{1} \times 0.80^{4}+10 \times 0.20^{2} \times 0.80^{3}+10 \times 0.20^{3} \times 0.80^{2}[/tex]
= 0.9933