15 balls, so
P(R,R) = 3/15 * 2/14
P(B,B) = 5/15 * 4/14
P(G,G) = 7/15 * 6/14
P(2 same) = sum of the above
A bag contains 3 red balls,5 blue balls and 7 green balls two are picked without replacement, what is the probability that both are different colour, they are of same colour
3 answers
15 balls
first red = 3/15
second red = 2/14
both red = 3/15 * 2/14 = 1/5 * 1/7 = 1/35
first blue = 5/15 = 1/3
second blue 4/14 = 2/7
both blue = 2/21
first green = 7/15
second green = 6/14 = 3/7
both green = 7/15 * 3/7 = 1/5
now add 1/35 + 2/21 + 1/5 = about .0286 + .0952 + .2
first red = 3/15
second red = 2/14
both red = 3/15 * 2/14 = 1/5 * 1/7 = 1/35
first blue = 5/15 = 1/3
second blue 4/14 = 2/7
both blue = 2/21
first green = 7/15
second green = 6/14 = 3/7
both green = 7/15 * 3/7 = 1/5
now add 1/35 + 2/21 + 1/5 = about .0286 + .0952 + .2
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