Probability of Drawing Two Red Marbles

To find the probability of drawing two red marbles from the bag without replacement, follow these steps: 1. **Determine the total number of marbles in the bag:** \[ 6 \text{ (red)} + 7 \text{ (blue)} + 9 \text{ (white)} = 22 \text{ (total marbles)} \] 2. **Calculate the probability of drawing a red marble on the first draw:** \[ P(\text{First red}) = \frac{6}{22} = \frac{3}{11} \] 3. **Calculate the probability of drawing a red marble on the second draw, given that the first marble drawn was red:** - After drawing one red marble, there are now 5 red marbles left. - There are now a total of 21 marbles remaining. \[ P(\text{Second red} | \text{First red}) = \frac{5}{21} \] 4. **Calculate the overall probability of both events happening (drawing two red marbles):** \[ P(\text{Two red marbles}) = P(\text{First red}) \times P(\text{Second red} | \text{First red}) = \frac{3}{11} \times \frac{5}{21} \] 5. **Multiply the fractions:** \[ \frac{3}{11} \times \frac{5}{21} = \frac{3 \times 5}{11 \times 21} = \frac{15}{231} \] 6. **Simplify the fraction:** \[ \frac{15}{231} \] The greatest common divisor (GCD) of 15 and 231 is 3: \[ \frac{15 \div 3}{231 \div 3} = \frac{5}{77} \] So, the probability of drawing two red marbles, without replacement, is: \[ \frac{5}{77} \]