Question

a bag contains 3 red marbles, 7 blue marbles and 8 green marbles. if two marbles are drawn out of the bag (without replacement), what is the probability, to the nearest 10th of a percent, that both marbles drawn will be blue?

Explanation:

Step1: Calculate total marbles

The total number of marbles is $3 + 7+8=18$.

Step2: Calculate first - draw probability

The probability of drawing a blue marble on the first draw is $\frac{7}{18}$.

Step3: Calculate second - draw probability

Since we don't replace, for the second draw, there are $17$ marbles left and $6$ blue marbles left. So the probability of drawing a blue marble on the second draw given that the first one was blue is $\frac{6}{17}$.

Step4: Calculate combined probability

The probability that both marbles are blue is the product of the two probabilities: $\frac{7}{18}\times\frac{6}{17}=\frac{7\times6}{18\times17}=\frac{42}{306}\approx0.13725$.

Step5: Convert to percentage

To convert to a percentage, we multiply by $100$: $0.13725\times 100 = 13.725\%$. Rounding to the nearest tenth of a percent gives $13.7\%$.

Answer:

$13.7\%$