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A bag contains 3 red marbles, 7 blue marbles and 4 green marbles. If three marbles are drawn out of the bag, what is the probability, to the nearest 10 th of a percent, that all three marbles drawn will be blue?
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Step 1 ：To find the probability that all three marbles drawn will be blue, we need to calculate the probability of drawing a blue marble on each of the three draws, and then multiply those probabilities together.

Step 2 ：Since there are 7 blue marbles out of a total of 14 marbles (3 red + 7 blue + 4 green), the probability of drawing a blue marble on the first draw is \( \frac{7}{14} \).

Step 3 ：After one blue marble is drawn, there will be 6 blue marbles left out of a total of 13 marbles, so the probability of drawing a blue marble on the second draw is \( \frac{6}{13} \).

Step 4 ：Finally, after two blue marbles are drawn, there will be 5 blue marbles left out of a total of 12 marbles, so the probability of drawing a blue marble on the third draw is \( \frac{5}{12} \).

Step 5 ：Multiplying these probabilities together will give the final answer: \( \frac{7}{14} \times \frac{6}{13} \times \frac{5}{12} \).

Step 6 ：The final probability is \( \frac{7}{14} \times \frac{6}{13} \times \frac{5}{12} = 0.09615384615384616 \).

Step 7 ：Rounding to the nearest 10th of a percent, the final probability is 9.6%.

Step 8 ：Final Answer: \( \boxed{9.6} \)