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A bag has 6 blue marbles, 4 red marbles, and 2 green marbles. Two marbles are randomly chosen without replacement. What is the probability that both marbles picked are blue? Answer options with 5 options A. StartFraction 1 over 6 EndFraction B. StartFraction 2 over 24 EndFraction C. StartFraction 5 over 22 EndFraction D. StartFraction 1 over 4 EndFraction E. StartFraction 1 over 3 EndFraction

Question

A bag has 6 blue marbles, 4 red marbles, and 2 green marbles. Two marbles are randomly chosen without replacement.

What is the probability that both marbles picked are blue?

Answer options with 5 options A. StartFraction 1 over 6 EndFraction

B. StartFraction 2 over 24 EndFraction

C. StartFraction 5 over 22 EndFraction

D. StartFraction 1 over 4 EndFraction

E. StartFraction 1 over 3 EndFraction

...expand
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Solution

To solve this problem, we need to calculate the probability of drawing a blue marble twice in a row from the bag.

Step 1: Calculate the total number of marbles in the bag. This is 6 (blue) + 4 (red) + 2 (green) = 12 marbles.

Step 2: Calculate the probability of drawing a blue marble on the first draw. This is the number of blue marbles (6) divided by the total number of marbles (12). So, the probability is 6/12 = 1/2.

Step 3: Now, we need to calculate the probability of drawing a blue marble on the second draw. But, since we've already drawn one blue marble and haven't replaced it, there are now only 5 blue marbles left, and the total number of marbles is 11. So, the probability is 5/11.

Step 4: The probability of both events happening (drawing a blue marble twice in a row) is the product of the probabilities of each event. So, we multiply the probabilities from Step 2 and Step 3 together: (1/2) * (5/11) = 5/22.

So, the correct answer is C. StartFraction 5 over 22 EndFraction.

This problem has been solved

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