Answer :
Using the binomial distribution, we have that:
a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.
b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.
c) The expected number of people is 4, with a variance of 20.
For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- p is the probability of a success on a single trial.
- n is the number of trials.
- p is the probability of a success on a single trial.
The expected number of trials before q successes is given by:
[tex]E = \frac{q(1-p)}{p}[/tex]
The variance is:
[tex]V = \frac{q(1-p)}{p^2}[/tex]
In this problem, 0.2 probability of a finding a person who attended the last football game, thus [tex]p = 0.2[/tex].
Item a:
- None of the first three attended, which is P(X = 0) when n = 3.
- Fourth attended, with 0.2 probability.
Thus:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512[/tex]
[tex]0.2(0.512) = 0.1024[/tex]
0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.
Item b:
This is the probability that none of the first six went, which is P(X = 0) when n = 6.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621[/tex]
0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.
Item c:
- One person, thus [tex]q = 1[/tex].
The expected value is:
[tex]E = \frac{q(1-p)}{p} = \frac{0.8}{0.2} = 4[/tex]
The variance is:
[tex]V = \frac{0.8}{0.04} = 20[/tex]
The expected number of people is 4, with a variance of 20.
A similar problem is given at https://brainly.com/question/24756209