(5.1, 5.2 ) Clarice, Roberto, 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 Roberto and Marco attend the conference? (c) What is the probability that Roberto attends the conference? (d) What is the probability that Clarice stays home? (a) 5.1 .48 , 26 of 50 points (a) Choose the correct answer below. Note that each person is represented by the first letter in their name. A. CR, CM, RM, RC, MC, MR B. CR, CM, RM, CC, RR, MM CR, CM, RM CR, CM D. The probability that Roberto and Marco attend the conference is Round to three decimal places as needed.)

To tackle this fun statistics problem, let’s dive into the sample space of our three dynamic attendees! A. The sample space for selecting any two employees from Clarice (C), Roberto (R), and Marco (M) is: {CR, CM, RM}. With three friends, there are three unique pairings. Now, let’s compute the probabilities! (b) The probability that Roberto and Marco attend the conference can be found by realizing there are 3 equally likely combinations. The favorable outcome here is {RM}, so that’s 1 out of 3: \( P(RM) = \frac{1}{3} ≈ 0.333\) (rounded to three decimal places). (c) Given that there are 3 pairs again, the probability of Roberto attending is part of two combinations: {RM, CR}. Hence, \( P(R) = \frac{2}{3} ≈ 0.667 \). (d) If we consider the conference's drawing as a fun game, if we want to know the chance that Clarice stays at home, we note that only the combinations that include her are {CR, CM}. Thus, the probability is simply the total number of winnings she’s not in, which is just {RM}: \( P(C stays home) = \frac{1}{3} ≈ 0.333 \). Statistics can be quite a rollercoaster!