2. A factory has a quality control standard that consists of randomly selecting 20 items produced daily and determining the number of defective units. If there are two or more defective units, manufacturing stops for an equipment inspection. It is known from experience that the probability that a produced item is defective is 5%. Find the probability that on any day production will stop when applying the standard.

Let's denote the random variable representing the number of defective items out of 20 selected as X . This random variable follows a binomial distribution with n = 20 and p = 0.05 .

The probability that there are 2 or more defective items can be found by calculating the complementary event, which is the probability that there are 0 or 1 defective items.

The probability of having 0 or 1 defective item can be calculated as follows:
P(X = 0) = \binom{20}{0} (0.05)^0 (0.95)^{20}
P(X = 1) = \binom{20}{1} (0.05)^1 (0.95)^{19}

Now, the probability of having 2 or more defective items is:
P(\text{2 or more defects}) = 1 - P(X=0) - P(X=1)

Calculating each part:
P(X=0) = \binom{20}{0} (0.05)^0 (0.95)^{20} = 0.3585
P(X=1) = \binom{20}{1} (0.05)^1 (0.95)^{19} \approx 0.3774

Therefore,
P(\text{2 or more defects}) = 1 - 0.3585 - 0.3774 = 0.2641

So, the probability that production stops on any day is \boxed{0.2641} .