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= 5
P(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 ≤ 5
P(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 ≥ 6
P(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)
Answer: a= 0.101 b= 0.158 c= 0.842
Step-by-step explanation: using the formula: nCr (pi)^r (1-pi)^n-r
