Answer :

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) = _{n}C_{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.**

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