Answer: 1) 0.6561 2) 0.0037
Step-by-step explanation:
We use Binomial distribution here , where the probability of getting x success in n trials is given by :-
[tex]P(X=x)=^nC_xp^x(1-p)^{n-x}[/tex]
, where p =Probability of getting success in each trial.
As per given , we have
The probability that any satellite dish owners subscribe to at least one premium movie channel. : p=0.10
Sample size : n= 4
Let x denotes the number of dish owners in the sample subscribes to at least one premium movie channel.
1) The probability that none of the dish owners in the sample subscribes to at least one premium movie channel = [tex]P(X=0)=^4C_0(0.10)^0(1-0.10)^{4}[/tex]
[tex]=(1)(0.90)^4=0.6561[/tex]
∴ The probability that none of the dish owners in the sample subscribes to at least one premium movie channel is 0.6561.
2) The probability that more than two dish owners in the sample subscribe to at least one premium movie channel.
= [tex]P(X>2)=1-P(X\leq2)\\\\=1-[P(X=0)+P(X=1)+P(X=2)]\\\\= 1-[0.6561+^4C_1(0.10)^1(0.90)^{3}+^4C_2(0.10)^2(0.90)^{2}]\\\\=1-[0.6561+(4)(0.0729)+\dfrac{4!}{2!2!}(0.0081)]\\\\=1-[0.6561+0.2916+0.0486]\\\\=1-0.9963=0.0037[/tex]
∴ The probability that more than two dish owners in the sample subscribe to at least one premium movie channel is 0.0037.
