Question 1192269
Here's how to approach this problem:

**1. Probability of a Drop Passing Through One Grid:**

* The drop will pass through a grid if its center falls within a certain area.  Imagine a smaller square inside the 1x1 grid. The drop will pass if its center falls within this smaller square.
* The side length of this smaller square is 1 - 2 * radius = 1 - 2 * 0.09 = 0.82.
* The area of this smaller square is 0.82 * 0.82 = 0.6724.  This is the probability of a drop passing through *one* grid.

**2. Probability of a Drop Passing Through 'n' Grids:**

* The probability of a drop passing through *n* grids is 0.6724ⁿ (since the grids are independent).

**3. Probability of a Drop Collapsing (Not Passing Through):**

* The probability of a drop *not* passing through *n* grids (and thus collapsing) is 1 - 0.6724ⁿ.

**4. Finding the Minimum Number of Grids (n):**

* We want this probability to be at least 0.69:
    1 - 0.6724ⁿ ≥ 0.69
    0.6724ⁿ ≤ 0.31
    n * ln(0.6724) ≤ ln(0.31)
    n ≥ ln(0.31) / ln(0.6724)   (Note: The inequality sign flips because ln(0.6724) is negative)
    n ≥ 2.82
* Since n must be an integer, we need at least 3 grids.

**5. Probability of Droplets Remaining Unbroken (Passing Through):**

* **0.09 Radius Droplets:**
    * Probability of passing through 3 grids: 0.6724³ ≈ 0.306
    * Proportion of 0.09 radius droplets remaining unbroken: 0.56 * 0.306 ≈ 0.171

* **0.14 Radius Droplets:**
    * The "effective" side length for these droplets is 1 - 2 * 0.14 = 0.72.
    * The probability of one such droplet passing through a grid is 0.72² = 0.5184.
    * Probability of passing through 3 grids: 0.5184³ ≈ 0.139
    * Proportion of 0.14 radius droplets remaining unbroken: 0.44 * 0.139 ≈ 0.061

**6. Total Proportion Remaining Unbroken:**

* Total proportion = (Proportion of 0.09 radius unbroken) + (Proportion of 0.14 radius unbroken)
* Total proportion ≈ 0.171 + 0.061 ≈ 0.232

**Therefore:**

* At least 3 grids are needed.
* Approximately 23.2% of the droplets would remain unbroken.