To find the probability that all three marbles drawn will be red, we first need to determine the total number of marbles and the number of ways to draw three red marbles.
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Total number of marbles: \[ 6 \text{ (red)} + 3 \text{ (blue)} + 5 \text{ (green)} = 14 \text{ (total marbles)} \]
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Number of ways to choose 3 red marbles: Since there are 6 red marbles, the number of ways to choose 3 red marbles from these is given by the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \). \[ \binom{6}{3} = \frac{6!}{3! \cdot (6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
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Number of ways to choose any 3 marbles from total 14 marbles: Using the same combination formula, the number of ways to choose any 3 marbles from 14 is: \[ \binom{14}{3} = \frac{14!}{3! \cdot (14-3)!} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = \frac{2184}{6} = 364 \]
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Calculating the probability: The probability \( P \) that all three marbles drawn are red is the number of favorable outcomes divided by the total number of outcomes: \[ P(\text{all red}) = \frac{\text{Number of ways to choose 3 red}}{\text{Total ways to choose 3}} = \frac{20}{364} \]
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Simplifying the probability: To simplify \( \frac{20}{364} \), we find the greatest common divisor (GCD) of 20 and 364. \[ \text{GCD}(20, 364) = 4 \] Simplifying: \[ \frac{20 \div 4}{364 \div 4} = \frac{5}{91} \]
Thus, the exact probability that all three marbles drawn will be red is: \[ \boxed{\frac{5}{91}} \]