Assuming that X is a binomial random variable with n = 10 and p = 0.25, find the probability, P for each of the following values of X. (a) X = 5. (Give the answer to four decimal places.) P(X = 5) = (b) X = 2. (Give the answer to four decimal places.) P(X = 2) = (c) X = 1. (Give the answer to four decimal places.) P(X = 1) = (d) X = 9. (Give the answer to seven decimal places.) P(X = 9) =

[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where P(x) is the probability of getting success in x trials, n is total number of trials and p is probability of getting success in each trial.

Given : X is a binomial random variable with parameters :-

n = 10 and p = 0.25

[tex]\text{a) P(x=5)}= ^{10}C_{5}(0.25)^5(0.75)^5\\\\=252(0.25)^5(0.75)^5\approx0.0584[/tex]

[tex]\text{b) P(x=2)}= ^{10}C_{2}(0.25)^2(0.75)^8\\\\=45(0.25)^2(0.75)^8\approx0.2816[/tex]

[tex]\text{c) P(x=1)}= ^{10}C_{1}(0.25)^1(0.75)^9\\\\=(10)(0.25)^1(0.75)^9\approx0.1877[/tex]

[tex]\text{d) P(x=9)}= ^{10}C_{9}(0.25)^1(0.75)^9\\\\=(10)0.25)^1(0.75)^9\approx0.1877117[/tex]