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) = nCx Px (1 – P)n – x
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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