SOLUTION: In a box containing 13 red balls, 9 green balls and 7 yellow balls. Two balls are drawn one by one without replacement, what is the probability that (correct your answers to 4 dec

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Question 1191825: In a box containing 13 red balls, 9 green balls and 7 yellow balls. Two balls are drawn one by one without replacement, what is the probability that
(correct your answers to 4 decimal places.)
(a) both of them are the same color?
(b) both of them are not green?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.
In a box containing 13 red balls, 9 green balls and 7 yellow balls.
Two balls are drawn one by one without replacement, what is the probability that
(correct your answers to 4 decimal places.)
(a) both of them are the same color?
(b) both of them are not green?
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In all, there are 13 + 9 + 7 = 29 balls. 

(a)  P = P(RR) + P(GG) + P(YY) =  = %2813%2A12+%2B+9%2A8+%2B+7%2A6%29%2F%2829%2A28%29 = 

       = 270%2F812 = 135%2F406 = 0.3325  (rounded).    ANSWER



(b)  P = %28%2813%2B7%29%2F29%29%2A%28%2813%2B7-1%29%2F28%29 = %2820%2F29%29%2A%2819%2F28%29 = 380%2F812 = 95%2F203 = 0.4680  (rounded).    ANSWER

Solved.

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If you are wondering why there is a difference between my answer for part (a) and that by the other tutor,
it is because there is an error in calculations by @greenestamps in P(GG).



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


(a) both the same color

This means (1st red AND 2nd red) OR (1st green AND 2nd green) OR (1st yellow AND 2nd yellow)

By basic concepts of probability, "AND" translates to a multiplication of the probabilities; "OR" translates to an addition of the probabilities. So

P(both red) = (13/29)(12/28) = 156/812
P(both green) = (8/29)(7/28) = 56/812
P(both yellow) = (7/29)(6/28) = 42/812
P(both the same color) = (156+56+42)/812 = 254/812

(b) both not green

P(both not green) = (20/29)(19/28) = 380/812

You can do the calculations and round to the required number of decimal places.