Example 6:
A company has a production line that produces 95% good products and 5% defective products. A sample of 100 products is taken from the production line. Find the probability that the sample will contain at most 2 defective products.

Solution:
Given,
p (probability of producing a defective product) = 0.05
q (probability of producing a good product) = 0.95
n (sample size) = 100
x (number of defective products) = 0, 1, or 2

We can use the Binomial probability formula to find the probability of at most 2 defective products in a sample of 100 products.

Binomial probability formula:
P(x) = nCx * p^x * q^(n-x)
where,
nCx = n! / (x! * (n-x)!)
p = probability of success
q = probability of failure (1-p)

We need to find P(x ≤ 2), which means we need to find the probability of having 0, 1, or 2 defective products in a sample of 100 products.

Calculation:
P(x ≤ 2) = P(x = 0) + P(x = 1) + P(x = 2)
= (100C0 * 0.05^0 * 0.95^100) + (100C1 * 0.05^1 * 0.95^99) + (100C2 * 0.05^2 * 0.95^98)
= (1 * 1 * 0.95^100) + (100 * 0.05 * 0.95^99) + (4950 * 0.0025 * 0.95^98)
= 0.6199 (approx)

Therefore, the probability that the sample will contain at most 2 defective products is 0.6199 (approx) or 61.99%.

Can u pls write the question as we can have different books also.