In a binomial distribution, n = 12 and π = .60.a. find the probability for x = 5? (round your answer to 3 decimal places.) probabilityb. find the probability for x ≤ 5? (round your answer to 3 decimal places.) probabilityc. find the probability for x ≥ 6?

For binomial distribute we use formula[tex]nCr (p)^r (q)^{n-r}[/tex]the probability mass function of the binomial distribution is:[tex]nCr (\pi)^r (1-\pi)^{n-r}[/tex]Given n = 12 and π = .6. (Replace the values)the probability for x = 5, we need to find P(x=5). Here r= 5P(x=5)= [tex]12C5(\pi )^5 (1-\pi)^{12-5}[/tex]= [tex]12C5(0.6 )^5 (0.4)^{7}[/tex]= 0.10090 = 0.101(b) the probability for x ≤ 5P(x<=5) = P(x=0) + P(x=1) +P(x=2)+ P(x=3) + P(x=4) + P(x=5)P(x<=5) = [tex]12C0(0.6 )^0(0.4)^{12}[/tex] + [tex]12C1(0.6 )^1(0.4)^{11}[/tex] +  [tex]12C2(0.6 )^2(0.4)^{10}[/tex] + [tex]12C3(0.6 )^3(0.4)^{9}[/tex] + [tex]12C4(0.6 )^4(0.4)^{8}[/tex] + [tex]12C5(0.6 )^5 (0.4)^{7}[/tex]= 0.15821229 = 0.158(c) find the probability for x ≥ 6P(x>=6) =P(x=6) + P(x=7) +P(x=8)+ P(x=9) + P(x=10) + P(x=11)+ P(x=12) P(x>=6) = [tex]12C6(0.6 )^6(0.4)^{6}[/tex] + [tex]12C7(0.6 )^7(0.4)^{5}[/tex] + [tex]12C8(0.6 )^8(0.4)^{4}[/tex] + [tex]12C9(0.6 )^9(0.4)^{3}[/tex] + [tex]12C10(0.6 )^{10}(0.4)^{2}[/tex] +   [tex]12C11(0.6 )^{11}(0.4)^{1}[/tex]  +  [tex]12C12(0.6 )^{12}(0.4)^{0}[/tex] = 0.8417877 = 0.0842 (rounded to 3 decimal places)