a has to be either 1 or 2 as multiplying by 4 keeps the number 4 digits.
(if a= 3 then quadrupling that number would be >12000 (5 digits)  )

if a was 1 then d would have to be 4 so our number abcd would be:
1bc4 and its reverse would be 4cb1 which cannot be as an integer multiplied by 4 has to be even.

Therefore a has to be 2.
Making d = 8.

When multiplying 2bc4 x 4  to be 8cb2, b and c cannot be too large as multiplying by 4 would then make the thousands place >8.

For example if b = 3 then even the lowest # in the 2300s,  2300 x 4 = 9200  would be too large as a = 9.

So b must be 0, 1 or 2
b cannot be 0 as 20cd x 4 = dc02 and multiples of 4 always with the last two digits as a multiple of 4.
b cannot = 2 since a is already 2 so b must = 1!

c cannot be 1, 2, or 8 as they have already been used

So abcd is now 21c8 x 4 = 8c12
If c = 0, 2108 x 4 does not end in 12
If c = 3, 2138 x 4 does not end in 12
If c = 4, 2148 x 4 does not end in 12
If c = 5, 2158 x 4 does not end in 12
If c = 6, 2168 x 4 does not end in 12
If c = 7, 2178 x 4 DOES end in 12
If c = 9, 2198 x 4 does not end in 12
c cannot be 1, 2, or 8 as they have already been used

2178 x 4 = 8712

Answer

Scenario 1: 3.2% and Scenario 2: 89.3%

We need to compare the actual positives with the sum of the true positives and false positives.  In Scenario 1, if 1 in 1000 (.001) have it then 999 out of 1000 (.999) do not.

In Scenario 2, if 1 in 5 (.2) have it then 4 out of 5 (.8) do not.

A: 333

The clocks in the first row represent 9 + 9 + 3
The calculators in the second row are each 10 (also 1234 added together is 10)
Each lightbulb in the third row is 15 (15 + 15 – 15) = 15.
Which means each ray above the lightbulb =3
So in the last row:
9 + 9 x (12 + 12 + 12)
9 + 9(36)
9 + 324
333