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
1) \[\left( \dfrac{27}{20} \right){{\left( \dfrac{19}{20} \right)}^{7}}\]
2) \[\left( \dfrac{533}{400} \right){{\left( \dfrac{19}{20} \right)}^{6}}\]
3) \[\left( \dfrac{153}{20} \right){{\left( \dfrac{1}{20} \right)}^{7}}\]
4) \[\left( \dfrac{35}{16} \right){{\left( \dfrac{1}{20} \right)}^{6}}\]

Answer 1

Hint: In this question we have to use the concept of Binomial distribution. We know that in Binomial distribution probability distribution function is given by, \[P\left( X=x \right)={}^{n}{{C}_{x}}{{p}^{x}}{{q}^{n-x}}\] where \[n\] denotes no. of trials, \[p\] denotes probability of success and \[q\] denotes probability of failure. Also \[p\] and \[q\] satisfies the relation \[p+q=1\].


Complete step-by-step solution:
Now we have to find the probability that a sample of \[8\] items will contain less than \[2\] defective items and we have given that the items produced by a firm are supposed to contain \[5\%\] defective items.
Let us suppose that
\[\begin{align}
  & \Rightarrow X=\text{Number of defective items} \\
 & \Rightarrow p=\text{Probability of getting a defective items} \\
 & \Rightarrow q=1-p \\
\end{align}\]
As we have given that the items produced by the firm contains \[5\%\] defective items
\[\begin{align}
  & \Rightarrow p=5\% \\
 & \Rightarrow q=1-\left( 5\% \right)=95\% \\
\end{align}\]
Now we have to find the probability that a sample of \[8\] items will contain less than \[2\] defective item.
\[\Rightarrow P\left( X<2 \right)=P\left( X=0 \right)+P\left( X=1 \right)\cdots \cdots \cdots \left( i \right)\]
But, by using Binomial Distribution we can write,
\[\Rightarrow P\left( X=x \right)={}^{n}{{C}_{x}}{{p}^{x}}{{q}^{n-x}}\]
Here, we have
\[\Rightarrow n=8,p=5\%=\dfrac{5}{100}\text{ and }q=95\%=\dfrac{95}{100}\]
Thus by substituting the values equation \[\left( i \right)\] becomes,
\[\Rightarrow P\left( X<2 \right)=\left( {}^{8}{{C}_{0}}{{\left( \dfrac{5}{100} \right)}^{0}}{{\left( \dfrac{95}{100} \right)}^{8}} \right)+\left( {}^{8}{{C}_{1}}{{\left( \dfrac{5}{100} \right)}^{1}}{{\left( \dfrac{95}{100} \right)}^{7}} \right)\]
As we know that, \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
\[\Rightarrow {}^{8}{{C}_{0}}=1,{}^{8}{{C}_{1}}=8\]
Thus we can write the above equation as
\[\Rightarrow P\left( X<2 \right)=\left( {{\left( \dfrac{5}{100} \right)}^{0}}{{\left( \dfrac{95}{100} \right)}^{8}} \right)+\left( 8{{\left( \dfrac{5}{100} \right)}^{1}}{{\left( \dfrac{95}{100} \right)}^{7}} \right)\]
By simplifying we get
\[\begin{align}
  & \Rightarrow P\left( X<2 \right)=\left( {{\left( \dfrac{19}{20} \right)}^{8}} \right)+\left( 8{{\left( \dfrac{1}{20} \right)}^{1}}{{\left( \dfrac{19}{20} \right)}^{7}} \right) \\
 & \Rightarrow P\left( X<2 \right)=\dfrac{{{19}^{8}}}{{{20}^{8}}}+8\left( \dfrac{{{19}^{7}}}{{{20}^{8}}} \right) \\
 & \Rightarrow P\left( X<2 \right)=\left( \dfrac{{{19}^{7}}}{{{20}^{8}}} \right)\left[ 19+8 \right] \\
 & \Rightarrow P\left( X<2 \right)=\left( \dfrac{{{19}^{7}}}{{{20}^{8}}} \right)\left( 27 \right) \\
 & \Rightarrow P\left( X<2 \right)=\left( \dfrac{{{19}^{7}}}{{{20}^{7}}} \right)\left( \dfrac{27}{20} \right) \\
 & \Rightarrow P\left( X<2 \right)=\left( \dfrac{27}{20} \right){{\left( \dfrac{19}{20} \right)}^{7}} \\
\end{align}\]
Hence the probability that a sample of \[8\] items will contain less than \[2\] defective item is \[\left( \dfrac{27}{20} \right){{\left( \dfrac{19}{20} \right)}^{7}}\].
Thus, option (1) is the correct option.

Note:In this type of question students have to take care in the calculation part. During simplification they have to use different rules of indices. Also students have to remember that \[{{a}^{0}}=1,{}^{n}{{C}_{0}}=1,{}^{n}{{C}_{1}}=n\]. Students have to note the definition of factorials also as it is required in the calculation of \[{}^{n}{{C}_{x}}\]. Also students have to note the calculation \[1-5\%=1-\left( \dfrac{5}{100} \right)=\left( \dfrac{95}{100} \right)=95\%\]