Respuesta :
Answer:
a) 0.0954
b) 0.9045
c) 0.1601
Step-by-step explanation:
We are given the following information:
We treat first-year students having computers as a success.
P(first-year students have computers) = 54.3% = 0.543
Then the number of first year students follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 3
a) P(None have computers)
[tex]P(x =0) \\\\= \binom{3}{0}(0.543)^0(1-0.543)^3\\\\= 0.0954[/tex]
b) P(At least one has a computer)
[tex]P(x \geq 1) \\\\= \binom{3}{0}(0.543)^0(1-0.543)^3+\binom{3}{1}(0.543)^1(1-0.543)^2+\binom{3}{3}(0.543)^3(1-0.543)^0\\\\=0.3402+0.4042+0.1601= 0.9045[/tex]
c) P(All have computers)
[tex]P(x =3) \\\\= \binom{3}{3}(0.543)^3(1-0.543)^0\\\\= 0.1601[/tex]
Using the binomial distribution, it is found that there is a:
a) 0.0954 = 9.54% probability that none have computers.
b) 0.9046 = 90.46% probability that at least one has a computer.
c) 0.1601 = 16.01% probability that all have computers.
For each incoming first-year students, there are only two possible outcomes, either they have computers, or they do not. The probability of an student having computer is independent of any other student, hence, the binomial distribution is used to solve this question.
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:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 54.3% of incoming first-year students have computers, hence [tex]p = 0.543[/tex].
- 3 students are selected at random, hence [tex]n = 3[/tex]
Item a:
This probability is P(X = 0), hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{3,0}.(0.543)^{0}.(0.457)^{3} = 0.0954[/tex]
0.0954 = 9.54% probability that none have computers.
Item b:
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0954 = 0.9046[/tex]
0.9046 = 90.46% probability that at least one has a computer.
Item c:
This probability is P(X = 3), hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{3,3}.(0.543)^{3}.(0.457)^{0} = 0.1601[/tex]
0.1601 = 16.01% probability that all have computers.
To learn more about the binomial distribution, you can check https://brainly.com/question/24863377
