This problem can be solved using Bernoulli trials.

Here n = 5 (as we are drawing 5 pens only)

Success is defined when we get a defective pen.

Let p denotes the probability of success and q probability of failure.

∴ p = 10/100 = 0.1

And q = 1 – 0.1 = 0.9

As we need to find probability of getting at most 1 defective pen.

Let X be a random variable denoting the probability of getting r number of defective pens.

∴ P (drawing atmost 1 defective pen) = P(X = 0) + P(X = 1)

The binomial distribution formula is:

P(x) = _nC_x P^x (1 – P)^{n – x}

Where:

x = total number of “successes.”

P = probability of success on an individual trial

n = number of trials

⇒ P(X = 0) + P(X = 1) =

∴ P(drawing at most 1 defective pen) =

⇒ P(drawing at most 1 defective pen) =

Our answer matches with option D.

∴ Option (D) is the only correct choice.