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

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

& \Rightarrow X=\text{Number of defective items} \\\ & \Rightarrow p=\text{Probability of getting a defective items} \\\ & \Rightarrow q=1-p \\\ \end{aligned}$$ As we have given that the items produced by the firm contains $$5\%$$ defective items $$\begin{aligned} & \Rightarrow p=5\% \\\ & \Rightarrow q=1-\left( 5\% \right)=95\% \\\ \end{aligned}$$ Now we have to find the probability that a sample of $$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{aligned} & \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{aligned}$$ Hence the probability that a sample of $$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\%$$