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If you draw 4 MMs from a bag containing 4 red 4 yellow 4 green and 4 brown ones what is the probability that you will get 1 of each color?
Wiki User
∙ 14y ago
The first M&M picked is not restricted to one colour, so the probability of it being an acceptable M&M (one whose colour is not already taken) is 1. For the second M&M, there are 15 left in the bag and 3 of these are not acceptable - because we just chose the other one of that colour. So the probability of the second being acceptable is 12/15 = 0.8. For the third, we are down to only 8 acceptable M&Ms (two colours to choose from, four of each) but have 14 left in the bag (two have been removed and presumably eaten already). This is 8/14 = 4/7 ~ 0.57. For the last pick, there are only four acceptable M&Ms left in the bag of 13; 4/13 is about 0.31 or 31%.
Multiplying these together, we see that the probability of having one of each of four colours is equal to 0.14065934065934065934065934065934 or 0.141 to 3 places. This is the same as 14.1% in total.
Wiki User
∙ 14y ago
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