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It is given that a shipment is accepted if only one or none of the tablets in a sample of 40 tablets is defective.

Let Xdenote the number of defective tablets.

Success is defined as getting a defective tablet.

The probability of success is computed below:

$$\begin{gathered} p=3\% \\ =\frac{3}{100} \\ =0.03 \end{gathered}$$

The probability of failure is computed below:

$$\begin{gathered} q=1-p \\ =1-0.03 \\ =0.07 \end{gathered}$$

The number of trials (n) is equal to 40.

The binomial probability formula used to compute the given probability is as follows:

$P\left(X=x\right)={}^n{C}_x{\left(p\right)}^x{\left(q\right)}^{n-x}$

Using the binomial probability formula, the probability that one or none of the tablets are defective is computed below:

$$\begin{gathered} P\left(X=0\right)+P\left(X=1\right)={}^{40}{C}_0{\left(0.03\right)}^0{\left(0.07\right)}^{40-0}+{}^{40}{C}_1{\left(0.03\right)}^1{\left(0.07\right)}^{40-1} \\ =0.2957+0.3658 \\ =0.6615 \\ 鈮�0.662 \end{gathered}$$

Thus, the probability that the shipment will be accepted is equal to 0.662.

The probability of 0.662 suggests that 66% of such shipments will be accepted, and 33% of the shipments will be rejected.

Since the probability of rejection of shipment is high, the supplier should consider improving the quality of tablets.