SOLUTION: I can't really figure this task out Dimensions of wire mesh squares are 1 x 1. How many at least such grids need to be taken so that when they are accidentally stacked on top of e

Question 1192269: I can't really figure this task out
Dimensions of wire mesh squares are 1 x 1. How many at least such grids need to be taken so that when they are accidentally stacked on top of each other, a drop of water with a radius of 0.09 falling from the top will collapse into a bundle of grids with a probability of at least 0.69? We assume that the drop will fall if it bounces off the wire of at least one grid. If we add this amount of grids and a stream of 56% droplets with radius of 0.09 and 44% droplets with radius of 0.14 falls into it, what proportion of the droplets would remain unbroken?
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
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.

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