There are 3 green cards , 2 red cards and 3 blue cards in a pile. Garry randomly draws 3 cards in a row. He does not replace a card after he draws it. What is the probability that Garry did not draw any green cards?

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:
\[
\boxed{\frac{5}{28}}.
\]