Question

a bag contains 3 red marbles, 2 blue marbles and 7 green marbles. if two marbles are drawn out of the bag (without replacement), what is the probability, to the nearest 1000th, 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 3 red, 2 blue, and 7 green marbles. So the total number of marbles \( n = 3 + 2 + 7 = 12 \).

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 \( P(\text{first red})=\frac{3}{12} \).

Step3: Probability of second red marble (without replacement)

After drawing one red marble, there are now 2 red marbles left and a total of 11 marbles left. So the probability of drawing a red marble second is \( P(\text{second red})=\frac{2}{11} \).

Step4: Probability of both red marbles

Since we want both events (first red and second red) to happen, we multiply the probabilities from Step 2 and Step 3. So \( P(\text{both red}) = \frac{3}{12}\times\frac{2}{11} \).
Simplify the fraction: \( \frac{3\times2}{12\times11}=\frac{6}{132}=\frac{1}{22}\approx0.045 \) (rounded to the nearest thousandth).

Answer:

\( 0.045 \)