Question

There are 6% defective items in a large bulk of items. The probability that a sample of 8 items will include not more than one defective item is \[1.42\times {{\left( 0.94 \right)}^{x}}\] . What is the value of x?

Answer 1

Hint: First of all, calculate the probability of defective items. Then, get the probability of non-defective items after subtracting the probability of defective items from 1. Here, we have a binomial distribution. We have the formula for the probability of defective items using binomial distribution is, \[^{8}{{C}_{d}}{{\left( 0.06 \right)}^{^{d}}}{{\left( 0.94 \right)}^{8-d}}\] , where d is the number of defective items. Now, using the binomial distribution, calculate the probability of 1 defective item and 0 defective items. Now, add these probabilities. Then, compare it with \[1.42\times {{\left( 0.94 \right)}^{x}}\] and solve it further to get the value of x.

Complete step-by-step answer:
According to the question, we have 6% defect items in a large bulk of items. The probability that a sample of 8 items will include not more than one defective item is \[1.42\times {{\left( 0.94 \right)}^{x}}\] .
The probability that a sample of 8 items will include not more than one defective item = \[1.42\times {{\left( 0.94 \right)}^{x}}\] …………………(1)
Out of all items, we have 6% defective items.
The probability of the defective items = \[6\%=\dfrac{6}{100}=0.06\] ………………………(2)
The probability of the non-defective items = \[1-0.06=0.94\] ………………………(3)
Let us assume that the number of defective items is d.
Out of all items we have only two possibilities that are defective items and non-defective items.
It means that the distribution is the binomial distribution.
Now, it is given that we have 8 items.
From equation (2), we have the probability of defective items.
From equation (3), we have the probability of non-defective items.
Using the binomial distribution, we can get the probability of defective items in a sample of 8 items.
\[^{8}{{C}_{d}}{{\left( \text{Probability of defective items} \right)}^{^{d}}}{{\left( \text{Probability of non-defective items} \right)}^{8-d}}\]
\[{{=}^{8}}{{C}_{d}}{{\left( 0.06 \right)}^{^{d}}}{{\left( 0.94 \right)}^{8-d}}\] …………………..(4)
We have to find the probability that a sample of 8 items will include not more than one defective item. Using equation (4), we can get the probability of defective items.
The number of defective items can be 0 or 1.
The Probability that the sample has 1 defective item = \[^{8}{{C}_{1}}{{\left( 0.06 \right)}^{^{1}}}{{\left( 0.94 \right)}^{8-1}}{{=}^{8}}{{C}_{1}}{{\left( 0.06 \right)}^{^{1}}}{{\left( 0.94 \right)}^{7}}\] ………………………………(5)
The Probability that the sample has 0 defective items = \[^{8}{{C}_{0}}{{\left( 0.06 \right)}^{^{0}}}{{\left( 0.94 \right)}^{8-0}}{{=}^{8}}{{C}_{0}}{{\left( 0.06 \right)}^{0}}{{\left( 0.94 \right)}^{8}}\] ………………………………(6)
The probability that a sample of 8 items will include not more than one defective item = \[\begin{align}
  & ^{8}{{C}_{1}}{{\left( 0.06 \right)}^{^{1}}}{{\left( 0.94 \right)}^{7}}{{+}^{8}}{{C}_{0}}{{\left( 0.06 \right)}^{0}}{{\left( 0.94 \right)}^{8}} \\
 & =8\left( 0.06 \right){{\left( 0.94 \right)}^{7}}+{{\left( 0.06 \right)}^{0}}{{\left( 0.94 \right)}^{8}} \\
 & ={{\left( 0.94 \right)}^{7}}\left\{ 8\left( 0.06 \right)+\left( 0.94 \right) \right\} \\
 & ={{\left( 0.94 \right)}^{7}}\left\{ \left( 0.48+0.94 \right) \right\} \\
\end{align}\]
\[={{\left( 0.94 \right)}^{7}}\times 1.42\] …………………..(7)
From equation (1), we have,
The probability that a sample of 8 items will include not more than one defective item = \[1.42\times {{\left( 0.94 \right)}^{x}}\]
Now, comparing equation (1) and equation (7), we get
\[\begin{align}
  & \Rightarrow {{\left( 0.94 \right)}^{7}}\times 1.42=1.42\times {{\left( 0.94 \right)}^{x}} \\
 & \Rightarrow {{\left( 0.94 \right)}^{7}}={{\left( 0.94 \right)}^{x}} \\
 & \Rightarrow 7=x \\
\end{align}\]
Here, we get \[x=7\] .
Therefore, the value of x is 7.

Note: In this question, one might get confused about why we have calculated the probability of 0 defected items. In the question, it is given that the probability of the number of defective items should not be more than 1. It means the maximum number of defective items is 1. There is no restriction on the minimum number of defective items. So, the number of defective items can be either 0 or 1.