Question 1200771: The dimensions of the square grids in the grid pattern are 1 × 1. What is the minimum number of such grids that need to be selected so that when they are randomly stacked on top of each other, a water droplet of radius r falling from the top would be absorbed by the surface with a probability of at least α? We assume that the droplet is absorbed if it hits at least one grid. If we stack this number of grids and a stream falls on it consisting of a% of r radius droplets and (100 - a)% of s radius droplets, what fraction of the grids would remain uncovered? ([α, r, s, a] = [0.72, 0.1, 0.13, 21])
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **1. Determine the Minimum Number of Grids for a Single Droplet**
* **Probability of Absorption by a Single Grid:**
* The droplet will be absorbed by a single grid if its center falls within a circle of radius r + 0.5 (since the grid size is 1x1 and the droplet has radius r).
* Area of the circle: π * (r + 0.5)²
* Area of the grid: 1 * 1 = 1
* Probability of absorption by a single grid: (π * (r + 0.5)²) / 1 = π * (r + 0.5)²
* **Probability of Absorption by 'n' Grids (Cumulative Probability):**
* Assuming independent events, the probability of the droplet not being absorbed by any of the 'n' grids is:
* (1 - Probability of absorption by a single grid)^n
* = (1 - π * (r + 0.5)²) ^ n
* Therefore, the probability of the droplet being absorbed by at least one of the 'n' grids is:
* 1 - (1 - π * (r + 0.5)²) ^ n
* **Find 'n' for Desired Probability (α):**
* 1 - (1 - π * (r + 0.5)²) ^ n ≥ α
* (1 - π * (r + 0.5)²) ^ n ≤ 1 - α
* n * log(1 - π * (r + 0.5)²) ≤ log(1 - α)
* n ≥ log(1 - α) / log(1 - π * (r + 0.5)²)
* **Substitute given values:**
* α = 0.72, r = 0.1
* n ≥ log(1 - 0.72) / log(1 - π * (0.1 + 0.5)²)
* n ≥ log(0.28) / log(1 - π * 0.36)
* n ≥ 2.04
* **Since 'n' must be an integer, the minimum number of grids is 3.**
**2. Calculate Fraction of Uncovered Grids for the Stream**
* **Probability of a droplet of radius 'r' being absorbed:** 0.72 (given)
* **Probability of a droplet of radius 's' being absorbed:**
* Calculate using the same formula as in step 1:
* 1 - (1 - π * (s + 0.5)²) ^ 3
* 1 - (1 - π * (0.13 + 0.5)²) ^ 3
* ≈ 0.836
* **Probability of a droplet NOT being absorbed:**
* Droplet of radius 'r': 1 - 0.72 = 0.28
* Droplet of radius 's': 1 - 0.836 = 0.164
* **Fraction of uncovered grids:**
* (0.21 * 0.28) + (0.79 * 0.164) ≈ 0.217
**Therefore:**
* **Minimum number of grids:** 3
* **Fraction of uncovered grids:** 0.217
**Note:**
* This calculation assumes that the droplets fall independently and uniformly on the grid surface.
* This analysis provides an estimate. Actual results may vary depending on the specific distribution of droplets within the stream.
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