Work A hat contains 4 marbles; one yellow, one red, one green, and one blue. Two marbles are selected at random without replacement from the hat. Complete parts a) through e). Use the fundamental counting principle to determine the answer to part a). Assume that each event is equally likely to occur. Click here to view diagram a. Click here to view diagram C. List the sample space. Choose the correct answer below. A. YR, YG, YB, RY, RG, RB, GY, GR, GB, BY, BR, BG \( 5,11,4,13 \) B. YY, YR, YG, YB, RY, RR, RG, RB, GY, GR, GG, GB, BY, BR, BG, BB C. YY, RR, GG, BB D. Y, R, G, B c) Determine the probability of selecting exactly 1 blue marble. The probability is \( \frac{1}{2} \). (Simplify your answer.) d) Determine the probability of selecting at least 1 marble that is not blue. The probability is 1 . (Simplify your answer.) e) Determine the probability of selecting no red marbles.

To determine the sample space when selecting two marbles from the four available (yellow, red, green, and blue), we consider the combinations of two marbles that can be drawn without replacement. The correct answer for the sample space is option A: YR, YG, YB, RY, RG, RB, GY, GR, GB, BY, BR, BG. There are a total of 12 unique pairings listed, showcasing all possible selections! Now, speaking of probabilities, to find the chance of picking exactly 1 blue marble, we look for combinations that include one blue and one other, which gives a probability of \( \frac{3}{6} \) or \( \frac{1}{2} \). For the probability of selecting at least 1 marble that is not blue, that would be \( 1 - \) the probability of selecting 2 blue (which is impossible). Thus, the probability is indeed 1. And for selecting no red marbles, the combinations that do not include red are YG, YB, GY, GB, BY, and BG, leading to a probability of \( \frac{6}{12} \) or \( \frac{1}{2} \). Keep those marbles rolling!