Clarice, Roberto, Dominique, John, and Marco 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 Dominique and John attend the conference? (c) What is the probability that Clarice attends the conference? (d) What is the probability that Roberto stays home? A. CR,CD,CJ,CM,RD,RJ,RM,DJ,DM,JM,RC,DC,JC,MC,DR,JR,MR,JD,MD,MJ (a) Choose the correct answer below. Note that each person is represented by the first letter in their name. B. CR, CD, CJ, CM, RD, RJ, RM, DJ, DM, JM C. CR, CD, CJ, CM, RD, RJ, RM, DJ, DM, JM, CC, RR, DD, JJ, MM D. CR, CD, CJ, CM

**(a)** There are 5 employees, and we choose 2, so the sample space consists of the \(\binom{5}{2} = 10\) possible pairs: \[ \{C,R\},\ \{C,D\},\ \{C,J\},\ \{C,M\},\ \{R,D\},\ \{R,J\},\ \{R,M\},\ \{D,J\},\ \{D,M\},\ \{J,M\}. \] In the answer options provided, this corresponds to **B. CR, CD, CJ, CM, RD, RJ, RM, DJ, DM, JM** --- **(b)** For Dominique and John to attend, the drawn pair must be \(\{D, J\}\). There is only 1 favorable outcome. Thus, the probability is \[ P(\{D, J\}) = \frac{1}{10}. \] --- **(c)** To find the probability that Clarice is one of the two attendees, note that if Clarice is selected, the other attendee is chosen from the remaining 4 employees. Thus, there are 4 favorable outcomes: \(\{C,R\}, \{C,D\}, \{C,J\},\) and \(\{C,M\}\). \[ P(\text{Clarice attends}) = \frac{4}{10} = \frac{2}{5}. \] --- **(d)** To find the probability that Roberto stays home, we consider all outcomes in which Roberto is not selected. With Roberto excluded, we are choosing 2 employees out of the remaining 4. There are \(\binom{4}{2} = 6\) favorable pairs. Hence, \[ P(\text{Roberto stays home}) = \frac{6}{10} = \frac{3}{5}. \]