Question

a bag contains 2 red marbles, 4 blue marbles and 5 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

Total marbles = 2 + 4+ 5 = 11

Step2: Calculate probability of first - blue marble

Probability of first blue marble = $\frac{4}{11}$

Step3: Calculate probability of second - blue marble

After drawing one blue marble, there are 3 blue marbles left and 10 marbles in total. Probability of second blue marble = $\frac{3}{10}$

Step4: Calculate probability of both blue

Probability of both blue = $\frac{4}{11}\times\frac{3}{10}=\frac{12}{110}\approx 0.1091$

Step5: Convert to percentage

$0.1091\times100 = 10.91\%$. Rounding to nearest tenth of a percent gives 10.9%

Answer:

10.9%