Question

An industry representative claims that 10 percent of all satellite dish owners subscribe to at least one premium movie channel. In an attempt to justify this claim, the representative will poll a randomly selected sample of dish owners.
Suppose that the representative's claim is true, and suppose that a sample of 4 dish owners is randomly selected. Assuming independence, use an appropriate formula to compute. (Do not round your intermediate calculation and round your answers to 4 decimal places.)

(1)The probability that none of the dish owners in the sample subscribes to at least one premium movie channel.
(2) The probability that more than two dish owners in the sample subscribe to at least one premium movie channel.

Answer 1

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 :-

$P(X=x)=^nC_xp^x(1-p)^{n-x}$

, 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 = $P(X=0)=^4C_0(0.10)^0(1-0.10)^{4}$

$=(1)(0.90)^4=0.6561$

∴ 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.

= $P(X>2)=1-P(X\leq2)\\\\=1-[P(X=0)+P(X=1)+P(X=2)]\\\\= 1-[0.6561+^4C_1(0.10)^1(0.90)^{3}+^4C_2(0.10)^2(0.90)^{2}]\\\\=1-[0.6561+(4)(0.0729)+\dfrac{4!}{2!2!}(0.0081)]\\\\=1-[0.6561+0.2916+0.0486]\\\\=1-0.9963=0.0037$

∴ The probability that more than two dish owners in the sample subscribe to at least one premium movie channel is 0.0037.