For this problem, the working equation would be the one usedfor repeated trials probability.
Probability = n!/r!(n - r)! * p^r * q(n - r)
The probability of success is p. If we treat the error as success, then p = 0.10 and consequently q= 1 - p = 0.9. The n is the number of trials equal to 20. Then, the r i the number of successes.
a. n=20, p = 0.10, q = 0.9, r = 5
Probability = 20!/5!(20 - 5)! * 0.10^5 * 0.9^(20-5) = 0.0319
b. n=20, p = 0.10, q = 0.9
For this part, you will have to add when r = 5, r=6, r=7 until r=20. That's a tedious process. So, it would be more reasonable to take r = 4, r = 3, r = 2, r = 1 and r = 0. Then, we will just find the difference.
For r=4,
P = 20!/4!(20 - 4)! * 0.10^4 * 0.9^(20-4) = 0.0898
For r=3,
P = 20!/3!(20 - 3)! * 0.10^3 * 0.9^(20-3) = 0.1901
For r=2,
P = 20!/2!(20 - 2)! * 0.10^2 * 0.9^(20-2) = 0.2852
For r=1,
P = 20!/1!(20 - 1)! * 0.10^1 * 0.9^(20-1) = 0.2702
For r=0,
P = 20!/0!(20 - 0)! * 0.10^0 * 0.9^(20-0) = 0.1216
Total P = 1 - (0.0898+0.1901+0.2852+0.2702+0.1216)
Total P = 0.2332