The scenario describes a binomial probability problem, where we are trying to determine the probability that a shipment of aspirin tablets will be accepted. In this case:

• The number of trials (tablets tested) $n = 2020$.
• The probability of success (a defective tablet) $p = 0.22$.
• The acceptance criteria is 0 or 1 defective tablet.

The probability of exactly $k$ defective tablets in $n$ trials follows the binomial distribution formula:

$P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$

We are interested in the probability that there are 0 or 1 defective tablet, which means we need to calculate:

$P(X = 0) + P(X = 1)$

Step 1: Calculate $P(X = 0)$

$P(X = 0) = \binom{2020}{0} \cdot (0.22)^0 \cdot (0.78)^{2020}$ $P(X = 0) = 1 \cdot 1 \cdot (0.78)^{2020}$

Step 2: Calculate $P(X = 1)$

$P(X = 1) = \binom{2020}{1} \cdot (0.22)^1 \cdot (0.78)^{2019}$ $P(X = 1) = 2020 \cdot 0.22 \cdot (0.78)^{2019}$

Let's calculate the values to find the total probability of acceptance.