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can u please help me with this problem?
2 balls are drawn successively without replacement from a box containing 4 red, 5 white, and 3 green balls. What is the probability of getting:
a. 2 balls of different colors
b. 2 balls with the same color
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The number of different ways to choose 2 balls from the 12 total is (12 choose 2) or:
Let's do part b first. The number of ways to choose 2 balls the same color is the sum of the ways to choose 2 red balls plus the number of ways to choose 2 white plus the number of ways to choose 2 green. The number of ways to choose 2 red is (4 choose 2):
The number of ways to choose 2 white is (5 choose 2) which is equal to 10.
The number of ways to choose 2 green is (3 choose 2) which is equal to 3.
The sum of these possibilities is 19. So the probability of choosing 2 balls of the same color is 19 out of 66 or 19/66= 28.78%
So now part a is real easy. If they're not the same color, they must be different colors. We accounted for 19 of the 66 ways, so the other 47 ways must give you balls of different colors. So the probability is 47 out of 66 or 71.21%.
We can verify this number. The 3 possible ways to pick different color balls are: Red & white, red & green, and white & green. The number of ways to pick a red and white is equal to the number of different ways to pick a red ball multipled by the number of ways to pick a white ball. This is (4 choose 1) times (5 choose 1) which is 4 * 5 = 20 . Likewise the number of ways to pick a red & green is (4 choose 1) * (3 choose 1) = 3 * 4 - 12. White and Green: 5 * 3 = 15. Add these 3 together: 20 + 12 + 15 = 47.
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