SOLUTION: Four black marbles and six white marbles are placed in a bowl. Three marbles are chosen at random, without replacement. What is the probability of choosing; at least one of each co

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Question 1191068: Four black marbles and six white marbles are placed in a bowl. Three marbles are chosen at random, without replacement. What is the probability of choosing; at least one of each colour?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


If you are just learning how to solve problems like this, it is good practice to find the probabilities of all possible outcomes and verify that the sum of the probabilities is 1.

Note when doing that it is easiest to NOT simplify the individual probabilities....

(1) 3 black, 0 white

%28C%284%2C3%29%2AC%286%2C0%29%29%2FC%2810%2C3%29+=+%284%2A1%29%2F120+=+4%2F120

(2) 2 black, 1 white

%28C%284%2C2%29%2AC%286%2C1%29%29%2FC%2810%2C3%29+=+%286%2A6%29%2F120+=+36%2F120

(3) 1 black, 2 white

%28C%284%2C1%29%2AC%286%2C2%29%29%2FC%2810%2C3%29+=+%284%2A15%29%2F120+=+60%2F120

(4) 0 black, 3 white

%28C%284%2C0%29%2AC%286%2C3%29%29%2FC%2810%2C3%29+=+%281%2A20%29%2F120+=+20%2F120

The sum of those probabilities is 120/120 = 1, giving us reassurance that we have done the calculations correctly.

Getting at least one of each color means cases (2) or (3); the combined probability of those two cases is 96/120 = 4/5.

ANSWER: 4/5