Answer with Step-by-step explanation:
We are given that
x=The number among six randomly selected household that have cable TV.
n=6, p=0.7
a.We have to calculate p(3)=p(X=3)
Binomial distribution formula :[tex]P(X=x)=nC_xp^xq^{n-x}[/tex]
q=1-p=1-0.7=0.3
[tex]P(X=3)=6C_3(0.7)^3(0.3)^3=\frac{6!}{3!3!}(0.7)^3(0.3)^3[/tex]
[tex]P(X=3)=\frac{6\times 5\times 4\times 3!}{3!3\times 2\times 1}(0.7)^3(0.3)^3=0.1852[/tex] ([tex]nC_r=\frac{n!}{r!(n-r)!}[/tex])
P(X=3)=0.1852
b.We have to calculate p(6)
[tex]P(X=6)=6C_6(0.7)^6=0.1176[/tex]
[tex]P(X=6)=0.1176[/tex]
Hence, the probability that all six household have cable TV=0.1176
c P(X=5)=[tex]6C_5(0.7)^5(0.3)[/tex]
[tex]P(X=5)=\frac{6!}{5!}(0.7)^5(0.3)=\frac{6\times 5!}{5!}(0.7)^5(0.3)=0.3025[/tex]
Hence, P(X=5)=0.3025
