A bag contains 5 red marbles, 7 blue marbles and 3 green marbles. If two marbles are drawn out of the bag, what is the probability, to the nearest 1oth of a percent, that both marbles drawn will be blue? Answer:

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Determine Total Marbles: Determine the total number of marbles in the bag. Add the number of red, blue, and green marbles together. 5 red + 7 blue + 3 green = 15 total marbles.Calculate First Blue Probability: Calculate the probability of drawing the first blue marble. Since there are 7 blue marbles out of 15 total marbles, the probability of drawing a blue marble first is 7/15.Calculate Second Blue Probability: Calculate the probability of drawing a second blue marble after the first one has been drawn. Now there are 6 blue marbles left and only 14 marbles total. The probability of drawing a second blue marble is 6/14.Multiply Probabilities: Multiply the probabilities from Step 2 and Step 3 to find the probability of both events occurring. (7/15) * (6/14) = 42/210 = 1/5.Convert to Percentage: Convert the probability into a percentage. 1/5 = 0.2 and as a percentage, this is 20%.Round to Nearest Tenth: Round the percentage to the nearest tenth of a percent. 20% is already at the nearest tenth of a percent, so no further action is needed.

A bag contains $5$ red marbles, $7$ blue marbles and $3$ green marbles. If two marbles are drawn out of the bag, what is the probability, to the nearest $1$ oth of a percent, that both marbles drawn will be blue? $\newline$ Answer:

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A bag contains $5$ red marbles, $7$ blue marbles and $3$ green marbles. If two marbles are drawn out of the bag, what is the probability, to the nearest $1$ oth of a percent, that both marbles drawn will be blue? $\newline$ Answer: