Q. 935.0( 1 Vote )

A box has 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement at most one is defective?
A.

B.

C.

D.

Correct

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) = nCx Px (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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