<u>Answer:</u>
<u>Step-by-step explanation:</u>
<u>Write the product:</u>
<u>Convert it to improper fraction:</u>
<u>Multiply:</u>
<u>Simplify:</u>
Hence, <u>the product is 21/32.</u>
Hoped this helped.
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:
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:
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:
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