Consider a monochrome computer display of 2048 by 1920 pixel

Consider a monochrome computer display of 2048 by 1920 pixels. A uniform areal random point generator selects pixels and lights them up if they are off and leaves them on if they are already on. Starting with all pixels off in each case, calculate the average fraction of pixels lit up after (a) 4.0 million draws, and (b) eight million draws, (d) In addition, calculate how many draws are necessary before there is on the average only one pixel left off.

Solution

number of pixels N = 2048*1920 = 3932160

In the first draw 1 -pixel will be lit as all are off

with the second draw the probability that the selected pixel already lit is 1/N

the probability that we select a pixel and it is lit is (1-1/N)

in the third draw he probability to select a pixel to lit is (1-2/N)

proceeding like this after r attempts the number of pixel that are lit

= 1+(1-1/N) +(1-2/N) + (1-3/N) + . . . . .. . . . . . . +(1-(r-1)/N)

= r -(1+2+3+  .. . . . . . . . .+(r-1))/N

the Neumerator the second term is an arthemetic series

      = (r-1)*r/2

Numbr of pixel lit after r attempts

     = r- (r-1)*r/2N

when we have 4 million attempts r = 4e+6

r being veery large we can approximate

n = r(1-r/2N)

number of pixels lit = 4e+6(1 - (4e+6)/2*3932160
                  = 1.965494

aftre 8 million draws = 8e+6(1-8e+6/2*3932160)

r is greater than 2N and hence all pixel will be lit.

to make all the pixel lit

we need to have
r(1-r/2N) =N