overall probability of several probabilities

If you're wondering if at least one of the five arguments proves the point, then the answer is 1 - Probability(none of them prove the point) which is
1 - (1 - 0.20)^5
= 1 - 0.32768
= 0.67232

Thanks markr.
But I do not understand the 2nd part of your answer.
It looks to me that if the probability that none of my five arguments proves the point is 0.67232, then the probability than one of my five arguments proves the point is 1 - 0.67232 = 0.32768 which would conflict with the 1st part of your answer.
Cordially. Bernard

You read it backwards. The probability of none of your arguments proving your point is (1 - 20%)^5 = .32768.

In my response, "which is" refers to the whole expression (1-P(...)), not just the P(...) part.

Thanks markr and engineer.
Cordially, Bernard

To markr and engineer,
Can you provide me with an equation if I have N independent arguments, but with each argument having a different probability (proving the same point) than the others.
Cordially, Bernard

1 - [(1-P1)*(1-P2)*...*(1-PN)]
where Pi is the probability that argument i proves the point.