Question

Suppose that in a large lot containingTmanufactured items, 30 percent of the items are defective, and 70 percent are non-defective. Also, suppose that ten items are selected randomly without replacement from the lot.

Determine (a) an exact expression for the probability that not more than one defective item will be obtained and (b) an approximate expression for this probability based on the binomial distribution.

Short answer

a) The probability that a maximum of one defective item will obtain is

\(\frac{{\left( {\begin{array}{{}{}}{0.7T}\\{10}\end{array}} \right) + 0.3T\left( {\begin{array}{{}{}}{0.7T}\\9\end{array}} \right)}}{{\left( {\begin{array}{{}{}}T\\{10}\end{array}} \right)}}\).

b) Based on the binomial distribution, the probability of obtaining a maximum of one defective item is \({\left( {0.7} \right)^{10}} + \left( {10 \times 0.3} \right) \times {\left( {0.7} \right)^9}\).

Step-by-step solution

Given information

There are T manufactured items in a large lot in which 30 percent of items are defective, and 70 percent of items are non-defective. The total number of items which are selected from the lot is 10.

Compute the probability

Let the probability that exactly x defective item will obtain as

\(P = \frac{{\left( {\begin{array}{{}{}}{0.3T}\\x\end{array}} \right)\left( {\begin{array}{{}{}}{0.7T}\\{10 - x}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}T\\{10}\end{array}} \right)}}\)

The probability that a maximum of one defective item will obtain is equal to the probability of obtaining 0 defective items and exactly one defective item.

So, the probability of obtaining a maximum of one defective item will obtain as

\(\begin{array}{c}{P_1} = \frac{{\left( {\begin{array}{}{0.3T}\\0\end{array}} \right)\left( {\begin{array}{}{0.7T}\\{10 - 0}\end{array}} \right)}}{{\left( {\begin{array}{}T\\{10}\end{array}} \right)}} + \frac{{\left( {\begin{array}{}{0.3T}\\1\end{array}} \right)\left( {\begin{array}{}{0.7T}\\{10 - 1}\end{array}} \right)}}{{\left( {\begin{array}{}T\\{10}\end{array}} \right)}}\\ = \frac{{\left( {\begin{array}{}{0.7T}\\{10}\end{array}} \right) + 0.3T\left( {\begin{array}{}{0.7T}\\9\end{array}} \right)}}{{\left( {\begin{array}{}T\\{10}\end{array}} \right)}}\end{array}\))

Hence, the probability that a maximum of one defective item will obtain is\(\frac{{\left( {\begin{array}{}{0.7T}\\{10}\end{array}} \right) + 0.3T\left( {\begin{array}{}{0.7T}\\9\end{array}} \right)}}{{\left( {\begin{array}{}T\\{10}\end{array}} \right)}}\).

(b) Compute the probability

Let the probability that exactly x defective item will obtain based on the binomial distribution, given as

\({P^ * } = \left( {\begin{array}{{}{}}{10}\\x\end{array}} \right){\left( {0.3} \right)^x}{\left( {0.7} \right)^{10 - x}}\)

So, based on the binomial distribution, the probability of obtaining a maximum of one defective item will obtain as:

\(\begin{array}{c}{P_2} = \left( {\begin{array}{{}{}}{10}\\0\end{array}} \right){\left( {0.3} \right)^0}{\left( {0.7} \right)^{10 - 0}} + \left( {\begin{array}{{}{}}{10}\\1\end{array}} \right){\left( {0.3} \right)^1}{\left( {0.7} \right)^{10 - 1}}\\ = {\left( {0.7} \right)^{10}} + \left( {10 \times 0.3} \right) \times {\left( {0.7} \right)^9}\end{array}\)

Hence, based on the binomial distribution, the probability of obtaining a maximum of one defective item will obtain is \({\left( {0.7} \right)^{10}} + \left( {10 \times 0.3} \right) \times {\left( {0.7} \right)^9}\).