Q. 935.0( 1 Vote )

# A box has 100 pen

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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