To find the probability that Garry did not draw any green cards when he randomly draws 3 cards from a pile of 3 green cards, 2 red cards, and 3 blue cards, we start by determining the total number of cards and the total ways to draw 3 cards.
Step 1: Total cards and combinations
The total number of cards is: \[ 3 \text{ (green)} + 2 \text{ (red)} + 3 \text{ (blue)} = 8 \text{ cards.} \] The total number of ways to draw 3 cards from 8 cards is calculated using the combinations formula \( \binom{n}{r} \): \[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56. \]
Step 2: Draw cards without green cards
Garry wants to draw cards without any green cards. The remaining cards are 2 red and 3 blue, totaling: \[ 2 \text{ (red)} + 3 \text{ (blue)} = 5 \text{ cards.} \] The number of ways to select 3 cards from these 5 cards is: \[ \binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10. \]
Step 3: Calculate the probability
The probability \( P \) that Garry does not draw any green cards is then the ratio of favorable outcomes to the total possible outcomes: \[ P(\text{no green cards}) = \frac{\text{Number of ways to choose 3 cards from 5 (non-green cards)}}{\text{Total ways to choose 3 cards from 8}} \] Substituting the values we found: \[ P(\text{no green cards}) = \frac{10}{56} = \frac{5}{28}. \]
Thus, the probability that Garry did not draw any green cards is: \[ \boxed{\frac{5}{28}}. \]