Question

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historically, the failure rate for LED light bulbs that the company manufactures is 5%. Suppose a random sample of 10 LED light bulbs is selected. What is the probability that a) None of the LED light bulbs are defective? b) Exactly one of the LED light bulbs is defective? c) Two or fewer of the LED light bulbs are defective? d) Three or more of the LED light bulbs are not defective?

Answer 1

Concepts and reason

The concept of binomial distribution is used to solve this problem.

A binomial experiment is a discrete probability experiment, which is repeated for a fixed number of trials and each of the trial is independent of the other trials. The possible outcomes for each trial are two and they are defined as success (S) or as a failure (F).

The probability of success, p for each trial remains the same. The random variable X in the experiment represents the number of successes in n independent trials.

Fundamentals

The probability mass function of binomial distribution is defined as,

$P\left( {x;p,n} \right) = \left( \begin{array}{l}\\n\\\\x\\\end{array} \right){p^x}{\left( {1 - p} \right)^{n - x}}$

Where, n is the number of trials, x is the number of success in n trials and p is the probability of success, and q is the probability of failures.

(a)

Consider the random variable X which represent the number of bulbs. The random variable X follows binomial distribution with $p = 0.05$ and $n = 10$ .

The probability that none of the LED light bulbs are defective.is calculated as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \left( \begin{array}{c}\\10\\\\0\\\end{array} \right){\left( {{\mkern 1mu} 0.05} \right)^0}{(1 - 0.05)^{10 - 0}}\\\\ = \frac{{10!}}{{0! \times \left( {10 - 0} \right)!}} \times {\left( {{\mkern 1mu} 0.05} \right)^0} \times {\left( {0.95} \right)^{10}}\\\\ = 0.5987\\\end{array}$

(b)

Consider the random variable X which represent the number of bulbs. The random variable X follows binomial distribution with $p = 0.05$ and $n = 10$ .

The probability that exactly one of the LED light bulbs is defective is calculated as,

$\begin{array}{c}\\P\left( {X = 1} \right) = \left( \begin{array}{c}\\10\\\\1\\\end{array} \right){\left( {{\mkern 1mu} 0.05} \right)^1}{(1 - 0.05)^{10 - 1}}\\\\ = \frac{{10!}}{{1! \times \left( {10 - 1} \right)!}} \times {\left( {{\mkern 1mu} 0.05} \right)^1} \times {\left( {0.95} \right)^9}\\\\ = 0.3151\\\end{array}$

(c)

Consider the random variable X which represent the number of bulbs. The random variable X follows binomial distribution with $p = 0.05$ and $n = 10$ .

The probability that two or fewer of the LED light bulbs are defective is calculated as,

$\begin{array}{c}\\P\left( {X \le 2} \right) = \sum\limits_{x = 0}^2 {\left( \begin{array}{c}\\10\\\\x\\\end{array} \right){{\left( {{\mkern 1mu} 0.05} \right)}^x}{{(1 - 0.05)}^{10 - x}}} \\\\ = \left( \begin{array}{l}\\\frac{{10!}}{{0! \times \left( {10 - 0} \right)!}} \times {\left( {{\mkern 1mu} 0.05} \right)^0} \times {\left( {0.95} \right)^{10}}\\\\ + \frac{{10!}}{{1! \times \left( {10 - 1} \right)!}} \times {\left( {{\mkern 1mu} 0.05} \right)^1} \times {\left( {0.95} \right)^9}\\\\ + \frac{{10!}}{{2! \times \left( {10 - 2} \right)!}} \times {\left( {{\mkern 1mu} 0.05} \right)^2} \times {\left( {0.95} \right)^8}\\\end{array} \right)\\\\ = 0.5987 + 0.3151 + 0.0746\\\\ = 0.9884\\\end{array}$

(d)

Consider the random variable X which represent the number of bulbs. The random variable X follows binomial distribution with $p = 0.05$ and $n = 10$ .

The probability that three or more of the LED light bulbs are defective is calculated as,

$\begin{array}{c}\\P\left( {X \ge 3} \right) = \sum\limits_{x = 3}^{10} {\left( \begin{array}{c}\\10\\\\x\\\end{array} \right){{\left( {{\mkern 1mu} 0.05} \right)}^x}{{(1 - 0.05)}^{10 - x}}} \\\\ = \left( \begin{array}{l}\\\frac{{10!}}{{3! \times \left( {10 - 3} \right)!}} \times {\left( {{\mkern 1mu} 0.05} \right)^3} \times {\left( {0.95} \right)^7} + \frac{{10!}}{{4! \times \left( {10 - 4} \right)!}} \times {\left( {{\mkern 1mu} 0.05} \right)^4} \times {\left( {0.95} \right)^6}\\\\ + \frac{{10!}}{{5! \times \left( {10 - 5} \right)!}} \times {\left( {{\mkern 1mu} 0.05} \right)^5} \times {\left( {0.95} \right)^5} + \frac{{10!}}{{6! \times \left( {10 - 6} \right)!}} \times {\left( {{\mkern 1mu} 0.05} \right)^6} \times {\left( {0.95} \right)^4}\\\\ + \frac{{10!}}{{7! \times \left( {10 - 7} \right)!}} \times {\left( {{\mkern 1mu} 0.05} \right)^7} \times {\left( {0.95} \right)^3} + \frac{{10!}}{{8! \times \left( {10 - 8} \right)!}} \times {\left( {{\mkern 1mu} 0.05} \right)^8} \times {\left( {0.95} \right)^2}\\\\ + \frac{{10!}}{{9! \times \left( {10 - 9} \right)!}} \times {\left( {{\mkern 1mu} 0.05} \right)^9} \times {\left( {0.95} \right)^1} + \frac{{10!}}{{10! \times \left( {10 - 10} \right)!}} \times {\left( {{\mkern 1mu} 0.05} \right)^{10}} \times {\left( {0.95} \right)^0}\\\end{array} \right)\\\\ = 0.0116\\\end{array}$

Ans: Part a

The probability that none of the LED light bulbs are defective.is 0.5987.

Part b

The probability that exactly one of the LED light bulbs is defective is 0.3151.

Part c

The probability that two or fewer of the LED light bulbs are defective is 0.9884.

Part d

The probability that three or more of the LED light bulbs are defective is 0.0116.