Question

4. At a local university 54.3% of incoming first-year students have computers. If 3 students are selected at random, and the following probabilities. a. None have computers. b. At least one has a computer. c. All have computers.

Answer 1

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

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

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)

$P(x =0) \\\\= \binom{3}{0}(0.543)^0(1-0.543)^3\\\\= 0.0954$

b) P(At least one has a computer)

$P(x \geq 1) \\\\= \binom{3}{0}(0.543)^0(1-0.543)^3+\binom{3}{1}(0.543)^1(1-0.543)^2+\binom{3}{3}(0.543)^3(1-0.543)^0\\\\=0.3402+0.4042+0.1601= 0.9045$

c) P(All have computers)

$P(x =3) \\\\= \binom{3}{3}(0.543)^3(1-0.543)^0\\\\= 0.1601$