A bag contains 6 red marbles, 8 blue marbles and 7 green marbles. If three marbles are dr: out of the bag, what is the exact probability that all three marbles drawn will be green?

To find the probability that all three marbles drawn will be green, we need to calculate the probability of drawing a green marble in each draw and then multiply these probabilities together. Given: - Total number of marbles in the bag = 6 red + 8 blue + 7 green = 21 marbles - Number of green marbles = 7 The probability of drawing a green marble in the first draw is the number of green marbles divided by the total number of marbles: \[ P(\text{green in first draw}) = \frac{7}{21} \] Since we are drawing without replacement, the probability of drawing a green marble in the second draw is the number of green marbles remaining divided by the total number of marbles remaining: \[ P(\text{green in second draw}) = \frac{6}{20} \] Similarly, the probability of drawing a green marble in the third draw is the number of green marbles remaining divided by the total number of marbles remaining: \[ P(\text{green in third draw}) = \frac{5}{19} \] The probability that all three marbles drawn will be green is the product of these probabilities: \[ P(\text{all green}) = P(\text{green in first draw}) \times P(\text{green in second draw}) \times P(\text{green in third draw}) \] Let's calculate this probability. Calculate the value by following steps: - step0: Calculate: \(\frac{\frac{\frac{7}{21}\times 6}{20}\times 5}{19}\) - step1: Reduce the fraction: \(\frac{\frac{\frac{1}{3}\times 6}{20}\times 5}{19}\) - step2: Reduce the fraction: \(\frac{\frac{1}{10}\times 5}{19}\) - step3: Multiply the numbers: \(\frac{\frac{1}{2}}{19}\) - step4: Multiply by the reciprocal: \(\frac{1}{2}\times \frac{1}{19}\) - step5: Multiply the fractions: \(\frac{1}{2\times 19}\) - step6: Multiply: \(\frac{1}{38}\) The probability that all three marbles drawn will be green is approximately 0.026316 or 1/38.