+182 A bag has 4 red marbles, 5 white marbles, and 6 blue marbles. Three marbles are drawn from the bag (without replacement). What is the probability that they are all the same color?
+1066 There are three cases to consider, since all three marbles can be red, all white, or all blue.
Case 1: All Red Marbles
Favorable cases: We need to choose 3 red marbles from 4 available. This can be done in ⁴C₃ ways (combinations, not permutations).
Total cases: We can choose any 3 marbles from a total of 4 red + 5 white + 6 blue = 15 marbles in ¹⁵C₃ ways.
Case 2: All White Marbles
Favorable cases: We need to choose 3 white marbles from 5 available. This can be done in ⁵C₃ ways.
Total cases: Same as Case 1 (¹⁵C₃).
Case 3: All Blue Marbles
Favorable cases: We need to choose 3 blue marbles from 6 available. This can be done in ⁶C₃ ways.
Total cases: Same as Case 1 (¹⁵C₃).
Total Probability
The probability that all three marbles are the same color (considering all three cases) is the sum of the probabilities of each case:
P(all same) = P(all red) + P(all white) + P(all blue)
Calculate Each Probability:
P(all red) = ⁴C₃ / ¹⁵C₃ P(all white) = ⁵C₃ / ¹⁵C₃ P(all blue) = ⁶C₃ / ¹⁵C₃
We can simplify each term using the fact that nCr = n! / (r! * (n-r)!):
P(all red) = (4!) / ((3!) * (1!)) / ((15!) / ((3!) * (12!))) P(all white) = (5!) / ((3!) * (2!)) / ((15!) / ((3!) * (12!))) P(all blue) = (6!) / ((3!) * (3!)) / ((15!) / ((3!) * (12!)))
Notice that (3!) and (12!) terms appear in all the denominators and numerators (except the first factorials in the numerators). These terms will cancel out, leaving:
P(all red) = 4 / (15 * 14 * 13) P(all white) = 10 / (15 * 14 * 13) P(all blue) = 20 / (15 * 14 * 13)
Final Probability:
Add the probabilities of each case:
P(all same) = 4/210 + 10/210 + 20/210 = 34/210
Simplify the fraction:
P(all same) = 17/105
Therefore, the probability that all three marbles drawn are the same color is 17/105.
