A box contains 2 red balls and 3 green balls. Two balls are drawn at random. What is the probability that both balls drawn are green?

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**Solution:**

1. Total number of balls = 2 red + 3 green = 5 balls.

2. Total number of ways to draw 2 balls out of 5 = $\binom{5}{2} = \frac{5 	imes 4}{2 	imes 1} = 10$.

3. Total number of balls given in the question is 7, but only 5 balls are described (2 red + 3 green). Assuming the total balls are 5 (as per the problem statement), the total possible outcomes for drawing 2 balls from 5 is 10.

4. Number of favorable outcomes (both balls green) = number of ways to choose 2 green balls out of 3 = $\binom{3}{2} = \frac{3 	imes 2}{2 	imes 1} = 3$.

5. Therefore, required probability = $\frac{	ext{favorable outcomes}}{	ext{total outcomes}} = \frac{3}{10}$.

**Note:** The original solution mentions total balls as 7 and favorable outcomes as 10, which seems inconsistent with the problem statement. Assuming the problem is to find the probability of drawing 2 balls from 7 balls where 2 are red and 3 are green, the remaining 2 balls are of unknown color, so the problem might be incomplete or misstated.

If the total balls are 7 with 2 red and 3 green, then the remaining 2 balls are not specified. If we consider only 5 balls (2 red + 3 green), then total outcomes = 10 and favorable outcomes = 3.

**Answer:** $\frac{3}{10}$

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**Alternatively, if the problem is exactly as given in the original solution:**

- Total balls = 7
- Number of ways to choose 2 balls = $\binom{7}{2} = 21$
- Number of favorable outcomes (drawing 2 balls out of 5 balls which are 2 red + 3 green) = $\binom{5}{2} = 10$

Then probability = $\frac{10}{21}$.

**Answer:** $\frac{10}{21}$

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