We define the random variable $X$ as the number of defective tablets in a sample of 22 tablets. Since each tablet is either defective or not, and the probability of a defective tablet is 2% (0.02), we model $X$ using the binomial distribution:

$X \sim \text{Binomial}(n=22, p=0.02)$

The shipment is accepted if at most 1 defective tablet is found, meaning we need to calculate:

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

Using the binomial probability mass function:

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

we calculate:

1. For $X = 0$: $P(X = 0) = \binom{22}{0} (0.02)^0 (0.98)^{22}$ $= (1)(1)(0.98)^{22}$

2. For $X = 1$: $P(X = 1) = \binom{22}{1} (0.02)^1 (0.98)^{21}$ $= 22 \cdot (0.02) \cdot (0.98)^{21}$

Now, we compute the values numerically. math import comb

Given values

n = 22 # sample size p = 0.02 # probability of defect q = 1 - p # probability of non-defect

Compute probabilities

P_X_0 = comb(n, 0) * (p ** 0) * (q ** 22) P_X_1 = comb(n, 1) * (p ** 1) * (q ** 21)

Total probability of acceptance

P_accept = P_X_0 + P_X_1 P_acceptThe probability that the whole shipment will be accepted is 0.9290 (rounded to four decimal places).

Would you like a more detailed explanation of the calculations? Here are some related questions:

1. How does the binomial distribution apply to quality control in pharmaceutical companies?
2. What happens to the acceptance probability if the defect rate increases to 5%?
3. How would this probability change if the sample size were increased to 30 tablets?
4. What is the probability of rejecting the shipment instead?
5. How could a company improve its sampling strategy to reduce the chance of accepting bad shipments?

Tip:

When using the binomial distribution for small defect probabilities, the Poisson approximation can sometimes simplify calculations!