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A bag contains 3 red marbles, 2 blue marbles and 4 green marbles. If two marbles are drawn out of the bag, what is the probability, to the nearest 10th of a percent, that both marbles drawn will be red?

User Changey
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2 Answers

5 votes

Answer:

1/9 ≅ 0.11

Step-by-step explanation:

User Conf
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3 votes
Answer: 8.3%

Step-by-step explanation:

To find the probability that both marbles drawn will be red, we need to find the probability of drawing a red marble on the first draw and then drawing another red marble on the second draw.

Total number of marbles in the bag = 3 red + 2 blue + 4 green = 9 marbles

Probability of drawing a red marble on the first draw:
P(Red₁) = (Number of red marbles) / (Total number of marbles) = 3/9 = 1/3

After drawing one red marble, there are now 2 red marbles left in the bag and 8 marbles total.

Probability of drawing a red marble on the second draw:
P(Red₂) = (Number of red marbles left) / (Total number of marbles left) = 2/8 = 1/4

Now, we need to find the probability of both events happening. We do this by multiplying the individual probabilities:

P(Red₁ and Red₂) = P(Red₁) × P(Red₂) = (1/3) × (1/4) = 1/12

Now let's convert this fraction to a percentage, rounded to the nearest 10th of a percent:

(1/12) × 100 ≈ 8.3%

So, the probability of drawing two red marbles consecutively is approximately 8.3%.
User Aracelis
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