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
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 <u>trials before q successes</u> is given by:
The variance is:
In this problem, 0.2 probability of a finding a person who attended the last football game, thus 
.
Item a:
- None of the first three attended, which is P(X = 0) when n = 3.
- Fourth attended, with 0.2 probability.
Thus:
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.
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
.
The expected value is:
The variance is:
The expected number of people is 4, with a variance of 20.
A similar problem is given at brainly.com/question/24756209
