4 ms·

Somethings off. It should work like this: p = 1e-6 n = 10^(1:7) frac_accepted = (1-p)^n Results: num_pixels frac_accepted 1e+01 0.99999000

by nonbel 9y ago

Somethings off. It should work like this: p = 1e-6 n = 10^(1:7) frac_accepted = (1-p)^n Results: num_pixels frac_accepted 1e+01 0.9999900000 1e+02 0.9999000049 1e+03 0.9990004993 1e+04 0.9900498288 1e+05 0.9048373728 1e+06 0.3678792572 1e+07 0.0000453997

Retric 9y ago
This assumes each failure is independent. Often a single failure will cause many bad pixels on the same screen. So you might have 98 good screens 1 with one bad pixel and one with 5,000 bad pixels when the per pixel odds say you are going to end up with 1 good screen if you are lucky and 99 bad ones.
- nonbel 9y ago
  I simulated 100 screens with different numbers of pixels where the probability the first pixel is bad in a screen is p0 = 1e-6 and the presence of each bad pixel increases the probability of the next one by the same amount (ie probability for a second bad pixel is 2e-6 and for last pixel is p0*number_already_bad_px). A screen was accepted if it had 0 bad pixels. The results look pretty much the same: num_pixels frac_accepted 1e+01 1.00 1e+02 1.00 1e+03 1.00 1e+04 0.98 1e+05 0.89 1e+06 0.32 1e+07 0.00 Of course this was just a convenient scheme to simulate. The greatest number of bad pixels was 86k for the 10 million px screen (so chance the last pixel was bad was ~9%). It does show that the independent pixel model isn't necessarily very far off though.
  - Retric 9y ago
    You messed up the math. You need to count (# number of pixels that where bad) / (total number of pixels produced) not simply call the odds for the first bad pixels as the rate of bad pixels.
    - nonbel 9y ago
      You are saying to adjust p0 so I get that the frequency of bad pixels = 1e-6? That makes sense but it would probably now depend on what size screens are being considered... not sure if I'll go back to it.
6nf 9y ago
I didn't actually calculate the probabilities, it was just my off-the-cuff estimates. 37% could be 'about half' :P

by nonbel 9y ago

Retric 9y ago

This assumes each failure is independent. Often a single failure will cause many bad pixels on the same screen. So you might have 98 good screens 1 with one bad pixel and one with 5,000 bad pixels when the per pixel odds say you are going to end up with 1 good screen if you are lucky and 99 bad ones.

nonbel 9y ago
I simulated 100 screens with different numbers of pixels where the probability the first pixel is bad in a screen is p0 = 1e-6 and the presence of each bad pixel increases the probability of the next one by the same amount (ie probability for a second bad pixel is 2e-6 and for last pixel is p0*number_already_bad_px). A screen was accepted if it had 0 bad pixels. The results look pretty much the same: num_pixels frac_accepted 1e+01 1.00 1e+02 1.00 1e+03 1.00 1e+04 0.98 1e+05 0.89 1e+06 0.32 1e+07 0.00 Of course this was just a convenient scheme to simulate. The greatest number of bad pixels was 86k for the 10 million px screen (so chance the last pixel was bad was ~9%). It does show that the independent pixel model isn't necessarily very far off though.
- Retric 9y ago
  You messed up the math. You need to count (# number of pixels that where bad) / (total number of pixels produced) not simply call the odds for the first bad pixels as the rate of bad pixels.
  - nonbel 9y ago
    You are saying to adjust p0 so I get that the frequency of bad pixels = 1e-6? That makes sense but it would probably now depend on what size screens are being considered... not sure if I'll go back to it.

6nf 9y ago

I didn't actually calculate the probabilities, it was just my off-the-cuff estimates. 37% could be 'about half' :P