Question

a bag contains 8 red marbles, 2 blue marbles and 4 green marbles. if two marbles are drawn out of the bag (without replacement), what is the probability, to the nearest tenth of a percent, that both marbles drawn will be red?

Explanation:

Step1: Find total number of marbles

First, we calculate the total number of marbles in the bag. There are 8 red, 2 blue, and 4 green marbles. So the total number of marbles \( n = 8 + 2 + 4 = 14 \).

Step2: Probability of first red marble

The probability of drawing a red marble first is the number of red marbles divided by the total number of marbles. So \( P(\text{first red})=\frac{8}{14} \).

Step3: Probability of second red marble (without replacement)

After drawing one red marble, there are now 7 red marbles left and a total of 13 marbles left. So the probability of drawing a second red marble is \( P(\text{second red}|\text{first red})=\frac{7}{13} \).

Step4: Probability of both red marbles

The probability of both events happening (drawing two red marbles without replacement) is the product of the two probabilities. So \( P(\text{both red}) = \frac{8}{14}\times\frac{7}{13} \).
Simplify the expression: \( \frac{8\times7}{14\times13}=\frac{56}{182}=\frac{4}{13}\approx0.3077 \).
To convert this to a percentage, we multiply by 100: \( 0.3077\times100 = 30.77\% \), which to the nearest tenth of a percent is \( 30.8\% \).

Answer:

\( 30.8\% \)