Question

A manufacturer of cotter pins knows that 5% of his product is defective. If he sells cotter pins in boxes of 100 and guarantees that not more than 10 pins will be defective, the approximate probability that a box will fail to meet the guaranteed quality is
(a) $\dfrac{{{e}^{-5}}{{5}^{10}}}{10!}$
(b) $1-\sum\limits_{x=0}^{10}{\dfrac{{{e}^{-5}}{{5}^{x}}}{x!}}$
(c) $1-\sum\limits_{x=0}^{\infty }{\dfrac{{{e}^{-5}}{{5}^{x}}}{x!}}$
(d) $\sum\limits_{x=0}^{\infty }{\dfrac{{{e}^{-5}}{{5}^{x}}}{x!}}$

Answer 1

Hint: Whenever you see not more than r more than terms in probability it will be maximum of Poisson’s distribution as the p is also small you can surely use Poisson’s distribution. First find the probability of a defective pin. By using this find the mean number of pins. Then by using these find probability for particular $x$ by Poisson’s distribution formula:

Complete step-by-step answer:

$P\left( X=x \right)=\dfrac{{{e}^{-m}}{{m}^{x}}}{x!}$

By above formula and given condition find the required probability

The total number of bones of cotton pins present:

n = 100

Let the probability of defective pins be the variable “p”

Given in the question 5% of his products are defective products. By using the above condition, we can write “p” as:

p = 5% = 0.05

Let the mean number of defective pins in a bar of 100 be “k”

Given in question: 5% of each bar are defective

By using the above condition, we can write “k” as

k = np = 5

Since p is small, we can use the Poisson’s distribution

By basic knowledge of distributions, probability in a Poisson’s distribution is written as:

$P\left( X=x \right)=\dfrac{{{e}^{-m}}{{m}^{x}}}{x!}$

By using the above condition, we have m = 5, we get:

$P\left( X=x \right)=\dfrac{{{e}^{-5}}{{5}^{x}}}{x!}$

We need the probability such that $x>10$ , in above equal.

Total probability is always 1. So, we can write it as:

$P\left( X>10 \right)+P\left( X\le 10 \right)=1$

By simplifying the above equation, $P\left( X>10 \right)$ can be written as

$P\left( X>10 \right)=1-P\left( X\le 10 \right)$

By using Poisson’s distribution here, we can write it as:

$P\left( X>10 \right)=1-\sum\limits_{x=0}^{10}{\dfrac{{{e}^{-5}}{{5}^{x}}}{x!}}$

This can be written as, by taking ${{e}^{-5}}$ term outside:

$P\left( X>10 \right)=1-{{e}^{-5}}\sum\limits_{x=0}^{10}{\dfrac{{{5}^{x}}}{x!}}$

This is the required probability of defective pins.

Therefore option(b) is the correct answer.

Note: Be careful while applying the Poisson’s distribution, don’t confuse $m,x$. $x$ is the value of a variable. m is the value of mean defective.