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  • 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?9/10^5

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. \left(\frac{9}{10}\right)^{5} \\\\B.\frac{1}{2}\left(\frac{9}{10}\right)^{4} \\\\C.. \frac{1}{2}\left(\frac{9}{10}\right)^{5} \\\\D. \left(\frac{9}{10}\right)^{5}+\frac{1}{2}\left(\frac{9}{10}\right)^{4}

Answers (1)

Solution

We can solve this using Bernoulli trials.

Here n = 5 (as we are drawing 5 pens only)
Success is defined as when we get a defective pen.
Let p be the probability of success and q be the probability of failure.
∴ p = 10/100 = 0.1
And q = 1 - 0.1 = 0.9
To find- the 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 at most 1 defective pen) = P(X = 0) + P(X = 1)
The binomial distribution formula is:

\mathrm{P}(\mathrm{x})=^{n} \mathrm{C}_{x} \mathrm{p}^{\mathrm{x}}(1-\mathrm{P})^{\mathrm{n}-\mathrm{x}}
Where:

x = total number of “successes.”
P = probability of success on an individual trial
n = number of trials

\Rightarrow P(X=0)+P(X=1)=5_{C_{0}} p^{0} q^{5}+{ }^{5} C_{1} p^{1} q^{4}

\mathrm{P}($ drawing at most 1 defective pen $)=\left(\frac{9}{10}\right)^{5}+5\left(\frac{1}{10}\right)\left(\frac{9}{10}\right)^{4}
\Rightarrow \mathrm{P}($ drawing at most 1 defective pen $) = \left ( \frac{9}{10} \right )^5+\frac{1}{2}\left ( \frac{9}{10} \right )^4
Our answer matches with option D.
∴ Option (D) is the only correct choice.

Posted by

infoexpert22

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