Questions: 2.) A box contains 12 balls identical in every respect except for color. There are 3 red balls, 7 green balls and 2 yellow balls. a.) Draw a tree diagram listing all the possibilities for your two selections. b.) Find the probability of each outcome if you select the two balls WITH replacement. c.) Find the probability of each outcome if you select the two balls WITHOUT replacement. d.) What is the probability of drawing two balls of the same color? (WITH replacement)

Solution Steps

a.) To draw a tree diagram, consider each selection as a branching point. For each ball drawn, list the possible colors and repeat for the second draw.

b.) With replacement, the probability of each outcome is calculated by multiplying the probability of drawing each ball on the first and second draw. Since the balls are replaced, the probabilities remain the same for each draw.

c.) Without replacement, the probability of each outcome is calculated by multiplying the probability of drawing each ball on the first draw and the adjusted probability on the second draw (since the first ball is not replaced, the total number of balls decreases by one).

When selecting two balls with replacement, the probabilities for each combination of colors are calculated as follows:

• \( P(RR) = P(R) \cdot P(R) = \left(\frac{3}{12}\right) \cdot \left(\frac{3}{12}\right) = 0.0625 \)
• \( P(RG) = P(R) \cdot P(G) = \left(\frac{3}{12}\right) \cdot \left(\frac{7}{12}\right) = 0.1458 \)
• \( P(RY) = P(R) \cdot P(Y) = \left(\frac{3}{12}\right) \cdot \left(\frac{2}{12}\right) = 0.0417 \)
• \( P(GR) = P(G) \cdot P(R) = \left(\frac{7}{12}\right) \cdot \left(\frac{3}{12}\right) = 0.1458 \)
• \( P(GG) = P(G) \cdot P(G) = \left(\frac{7}{12}\right) \cdot \left(\frac{7}{12}\right) = 0.3403 \)
• \( P(GY) = P(G) \cdot P(Y) = \left(\frac{7}{12}\right) \cdot \left(\frac{2}{12}\right) = 0.0972 \)
• \( P(YR) = P(Y) \cdot P(R) = \left(\frac{2}{12}\right) \cdot \left(\frac{3}{12}\right) = 0.0417 \)
• \( P(YG) = P(Y) \cdot P(G) = \left(\frac{2}{12}\right) \cdot \left(\frac{7}{12}\right) = 0.0972 \)
• \( P(YY) = P(Y) \cdot P(Y) = \left(\frac{2}{12}\right) \cdot \left(\frac{2}{12}\right) = 0.0278 \)

When selecting two balls without replacement, the probabilities for each combination of colors are calculated as follows:

• \( P(RR) = P(R) \cdot P(R|R) = \left(\frac{3}{12}\right) \cdot \left(\frac{2}{11}\right) = 0.0455 \)
• \( P(RG) = P(R) \cdot P(G|R) = \left(\frac{3}{12}\right) \cdot \left(\frac{7}{11}\right) = 0.1591 \)
• \( P(RY) = P(R) \cdot P(Y|R) = \left(\frac{3}{12}\right) \cdot \left(\frac{2}{11}\right) = 0.0455 \)
• \( P(GR) = P(G) \cdot P(R|G) = \left(\frac{7}{12}\right) \cdot \left(\frac{3}{11}\right) = 0.1591 \)
• \( P(GG) = P(G) \cdot P(G|G) = \left(\frac{7}{12}\right) \cdot \left(\frac{6}{11}\right) = 0.3182 \)
• \( P(GY) = P(G) \cdot P(Y|G) = \left(\frac{7}{12}\right) \cdot \left(\frac{2}{11}\right) = 0.1061 \)
• \( P(YR) = P(Y) \cdot P(R|Y) = \left(\frac{2}{12}\right) \cdot \left(\frac{3}{11}\right) = 0.0455 \)
• \( P(YG) = P(Y) \cdot P(G|Y) = \left(\frac{2}{12}\right) \cdot \left(\frac{7}{11}\right) = 0.1061 \)
• \( P(YY) = P(Y) \cdot P(Y|Y) = \left(\frac{2}{12}\right) \cdot \left(\frac{1}{11}\right) = 0.0152 \)

Final Answer