Find the following probabilities 1.P(Two marbles of the same color) 2.P(Two marbles of different colors) 3.P(red and blue in two marbles)?

OK we have n(r) = 2, n(b) = 2, and n(g) = 2 with SUM(n) = N = 6 marbles.

1. P(n(r,r) = 2 OR n(b,b) = 2 OR n(g,g) = 2)) = (1/3*1/5)*3 = .20 ANS. NOTE each pair of trials has the same likelihood; so we multiply by 3 for the three colors EX: First trial 2 colors out of six possibilities = 2/6 = 1/3. Second trial 1 color out of five = 1/5..

2. P(n(r,b) = 1 OR n(r,g) = 1 OR n(b,r) = 1 OR n(g,r) = 1 OR n(b,g) = 1 OR n(g,b)) = 1) = (1/3*2/5) * 6 = .80 ANS.Again each pair of trials has the same likelihood but there are six possible pairs; so we multiply by 6. Note the red then blue and blue then red make up two different trials.

3. P(n(r,b) = 1 OR n(b,r) = 1) = 1 - (4/6)*(4/5) = .47 ANS. Note: Number of ways first draw is not red or blue is 4 out of six marbles; number of ways second is not red or blue is also four, but now out of five remaining. So 4/6 * 4/5 is the probability of not drawing b then r or r then b. Which make 1 - (4/6)*(4/5) the probability of not not drawing or, ta da, the probability of drawing them. The negative of a negative is a positive.

NOTE: n(x,y) = 2 means the number of ways to draw x then y without replacement (e.g., 2 ways).