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A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas City, Kansas until he finds a person who attended the last home football game. Let p, the probability that he succeeds in finding such a person, equal 0.20. And, let X denote the number of people he selects until he finds his first success.
a. What is the probability that the marketing representative must select 4 people to find one who attended the last home football game?
b. What is the probability that the marketing representative must select more than 6 people to find one who attended the last home football game?
c. How many people should we expect (that is, what is the average number) the marketing representative needs to select before he finds one who attended the last home football game? And, while we're at it, what is the variance?

Sagot :

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

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