Asked by amelia
An urn contains 4 red, 6 white, and 3 blue marbles. Three marbles are selected at random and without replacement. What is the probability that one of each color is selected? Give your answer as a decimal number, with 3 decimal places.
Answers
Answered by
Damon
prob of red, white then blue
= 4/13 * 6/12 * 3/11
prob of red, blue then white
= 4/13 * 3/12 * 6/11
prob of white, red then blue
= 6/13 * 4/12 * 3/11
prob of white, blue then red
= 6/13 *3/12 * 4/11
prob of blue, red then white
=3/13 * 4/12 * 6/11
prob of blue, white then red
= 3/13 * 6/12 * 4/11
NOTE ---- those six ways are all the same probability
numerator = 3*6*4 = 72
denominator = 13 * 12 *11 = 1716
SO
6 (72/1716) = .252
= 4/13 * 6/12 * 3/11
prob of red, blue then white
= 4/13 * 3/12 * 6/11
prob of white, red then blue
= 6/13 * 4/12 * 3/11
prob of white, blue then red
= 6/13 *3/12 * 4/11
prob of blue, red then white
=3/13 * 4/12 * 6/11
prob of blue, white then red
= 3/13 * 6/12 * 4/11
NOTE ---- those six ways are all the same probability
numerator = 3*6*4 = 72
denominator = 13 * 12 *11 = 1716
SO
6 (72/1716) = .252
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