To solve this problem, we need to find the probability that both balls drawn are red when two balls are drawn at random from the bag without replacement.
Step 1: Calculate Total Number of Ways to Draw Two Balls
The bag contains 5 red, 3 blue, and 2 green balls, making a total of 10 balls.
The total number of ways to choose 2 balls out of 10 is given by the combination formula:
\(C(n, r) = \frac{n!}{r!(n-r)!}\)
Therefore, the number of ways to choose 2 balls from 10 is:
\(C(10, 2) = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45\)
Step 2: Calculate Number of Favorable Ways to Draw Two Red Balls
The number of ways to choose 2 red balls from the 5 red balls is:
\(C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10\)
Step 3: Calculate the Probability
The probability that both balls drawn are red is given by the ratio of the number of favorable outcomes to the total number of outcomes:
\(\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10}{45} = \frac{2}{9}\)
Therefore, the probability that both balls are red is \(\frac{2}{9}\).
Conclusion:
The correct answer is \(\frac{2}{9}\).