6. Since parts 4 and 5 say the same thing and P has not been defined as a statistical distribution, I'll assume a binomial distribution (binomial because there are only two states: defective or not defective), where p=80/100=0.8, q=0.2 and n=20. So the mean is np=16 and the variance is npq=3.2, and standard deviation is √(npq)=1.79 approx.
1. Exactly one defect in 20, means one defective CD and 19 perfect CDs.
The probability is calculated by starting with the probability of the first CD being defective=0.8 combined with the probability of the remainder being non-defective=0.219=5.24×10-14. But since the defective CD could be the first, or the second, or the third CD, ... there are 20 positions so we have:
20×0.8×5.24×10-14=8.39×10-13. This is a very tiny probability. We can expect such a low value because the defect rate is so high. 16 is the average (or mean) and 1 is so small compared to 16. In fact, this suggests that the question may be at fault, and perhaps only 20/100 CDs or 8/100 CDs are defective. The probabilities would then be more realistic.
If we use the normal distribution we have to calculate a suitable Z value, but because the sample size is only 20 we need to adjust the SD: 1.79/√20=0.4 approx. Also, we need two Z values because the normal distribution is continuous (not discrete), so we need two X values for each Z. We choose 0.5 and 1.5 so that 1 is sandwiched between. Z1=(0.5-16)/0.4=-38.75, Z2=(1.5-16)/0.4=-36.25. These correspond to extremely low probabilities.
4/5. If no more than two are defective we need to add the probabilities of none, 1 and 2. Again, whichever distribution we use, the probabilities are going to be very small. Let's look at the binomial distribution for two defective CDs out of 20. This time we have 190×0.82×0.218=3.19×10-11. This is still very small.
Even if the figures in the question are wrong, this solution should give some idea how to solve the problem with different statistics.