Question

a bag contains 7 red marbles, 8 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, 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 7 red, 8 blue, and 4 green marbles. So the total number of marbles \( n = 7 + 8 + 4 = 19 \).

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 \( P(\text{first red})=\frac{7}{19} \).

Step3: Probability of second red marble (without replacement)

After drawing one red marble, the number of red marbles left is \( 7 - 1 = 6 \) and the total number of marbles left is \( 19 - 1 = 18 \). So the probability of drawing a red marble second is \( P(\text{second red})=\frac{6}{18} \).

Step4: Probability of both red marbles

Since we want the probability of both events (first red and second red) happening, we multiply the two probabilities. So \( P(\text{both red}) = P(\text{first red}) \times P(\text{second red})=\frac{7}{19}\times\frac{6}{18} \).
Simplify the expression: \( \frac{7\times6}{19\times18}=\frac{42}{342}\approx0.12 \) (rounded to the nearest hundredth).

Answer:

\( 0.12 \)