Question

The items produced by a firm are supposed to contain $5\%$ defective items. The probability that a sample of $8$ items will contain less than $2$ defective item is

• A. $\left(\frac{27}{20}\right){\left(\frac{19}{20}\right)}^7$
• B. $\left(\frac{533}{400}\right){\left(\frac{19}{20}\right)}^6$
• C. $\left(\frac{153}{20}\right){\left(\frac{1}{20}\right)}^7$
• D. $\left(\frac{35}{16}\right){\left(\frac{1}{20}\right)}^6$

Answer 1

The correct option is A

$\left(\frac{27}{20}\right){\left(\frac{19}{20}\right)}^7$

Explanation for the correct option:

Find probability using the binomial probability theorem.

• Given: The probability of items supposed to have defected is $5\%$.
• Sample of $8$ items and $2$ of them are defective.
• Let the number of defective items be $x$. so, $x=2$.
• Let $P$ be the probability of getting a defective item and $P$ is $5\%$. So, $P=\frac{5}{100}$.

Assume that, $q$ is the probability of containing undefeated items.

We know that the sum of probability is $1$.

So,

$$\begin{gathered} q=1-p \\ \Rightarrow q=1-\frac{5}{100} \\ \Rightarrow q=\frac{95}{100} \end{gathered}$$

For $P(x<2)$ assume that $P_1$ be the probability that a sample of $8$ items will contain less than $2$ defective item.

$$\begin{gathered} P_1=P(x=0)+P(x=1) \\ \Rightarrow P_1={}^{8}C_0{\left(\frac{5}{100}\right)}^0{\left(\frac{95}{100}\right)}^8+{}^{8}C_1{\left(\frac{5}{100}\right)}^1{\left(\frac{95}{100}\right)}^7 \\ \Rightarrow P_1={}^{8}C_0{\left(\frac{1}{20}\right)}^0{\left(\frac{19}{20}\right)}^8+{}^{8}C_1{\left(\frac{1}{20}\right)}^1{\left(\frac{19}{20}\right)}^7 \\ \Rightarrow P_1={\left(\frac{19}{20}\right)}^8+8\times \left(\frac{1}{20}\right)\times {\left(\frac{19}{20}\right)}^7 \\ \Rightarrow P_1={\left[\frac{19}{20}\right]}^7\left(\frac{19}{20}+\frac{8}{20}\right) \\ \Rightarrow P_1={\left[\frac{19}{20}\right]}^7\left(\frac{27}{20}\right) \end{gathered}$$

Therefore, the required probability is ${\left[\frac{19}{20}\right]}^7\left(\frac{27}{20}\right)$.

Hence, the probability that a sample of $8$ items will contain less than $2$ defective item is ${\left[\frac{19}{20}\right]}^7\left(\frac{27}{20}\right)$.