Question

clance, roberto, and john work for a publishing company. the company wants to send two employees to a statistics conference. to be fair, the company decides that the two individuals who get to attend will have their names randomly drawn from a hat.
(a) determine the sample space of the experiment. that is, list all possible simple random samples of size n = 2
(b) what is the probability that clance and roberto attend the conference?
(c) what is the probability that clance attends the conference?
(d) what is the probability that roberto stays home?

(a) choose the correct answer below. note that each person is represented by the first letter in their name.
a. cr, cj
b. cr, cj, rj, rc, jc, jr
c. cr, cj, rj, cc, rr, jj
d. cr, cj, rj

(b) the probability that clance and roberto attend the conference is
(round to three decimal places as needed.)

Explanation:

Step1: List sample - space

The three employees are Clance (C), Roberto (R), and John (J). The sample space of choosing 2 - person samples from 3 people is all possible pairs. The pairs are CR, CJ, RJ.

Step2: Calculate probability for (b)

The probability of an event \(E\) is \(P(E)=\frac{n(E)}{n(S)}\), where \(n(E)\) is the number of elements in event \(E\) and \(n(S)\) is the number of elements in the sample - space \(S\). The event that Clance and Roberto attend the conference is a single outcome in the sample space. Since \(n(S) = 3\) (the three pairs CR, CJ, RJ) and \(n(E)=1\) (the pair CR), \(P=\frac{1}{3}\approx0.333\).

Step3: Calculate probability for (c)

The event that Clance attends the conference consists of the pairs CR and CJ. So \(n(E) = 2\) and \(n(S)=3\). Then \(P=\frac{2}{3}\approx0.667\).

Step4: Calculate probability for (d)

The event that Roberto stays home means the pairs are CJ. So \(n(E) = 1\) and \(n(S)=3\). Then \(P=\frac{1}{3}\approx0.333\).

Answer:

(a) D. CR, CJ, RJ
(b) 0.333
(c) 0.667
(d) 0.333