Manufacturing company

It is established by the production manager of a manufacturing company that 2 out of every 20 units of a particular product are faulty. In a particular production hour, 8 units of the product were produced. Determine the probability that:
(1) none of them is faulty
(2) At most four of them are faulty
(3) More than 3 of them are faulty
(4) exactly 2 of them are faulty

Correct answer:

a =  0.4305
b =  0.9996
c =  0.005
d =  0.1488

Step-by-step explanation:

$$\begin{gathered} n=8 \\ p=\frac{2}{20}=\frac{1}{10}=0.1 \\ {a}_0=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot p^0\cdot (1-p{)}^{n-0}=(\genfrac{}{}{0ex}{}{8}{0})\cdot 0.1^0\cdot (1-0.1{)}^{8-0}=1\cdot 0.1^0\cdot 0.9^8\doteq 0.4305 \\ {a}_1=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot p^1\cdot (1-p{)}^{n-1}=(\genfrac{}{}{0ex}{}{8}{1})\cdot 0.1^1\cdot (1-0.1{)}^{8-1}=8\cdot 0.1^1\cdot 0.9^7\doteq 0.3826 \\ {a}_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot p^2\cdot (1-p{)}^{n-2}=(\genfrac{}{}{0ex}{}{8}{2})\cdot 0.1^2\cdot (1-0.1{)}^{8-2}=28\cdot 0.1^2\cdot 0.9^6\doteq 0.1488 \\ {a}_3=\left(\genfrac{}{}{0ex}{}{n}{3}\right)\cdot p^3\cdot (1-p{)}^{n-3}=(\genfrac{}{}{0ex}{}{8}{3})\cdot 0.1^3\cdot (1-0.1{)}^{8-3}=56\cdot 0.1^3\cdot 0.9^5\doteq 0.0331 \\ {a}_4=\left(\genfrac{}{}{0ex}{}{n}{4}\right)\cdot p^4\cdot (1-p{)}^{n-4}=(\genfrac{}{}{0ex}{}{8}{4})\cdot 0.1^4\cdot (1-0.1{)}^{8-4}=70\cdot 0.1^4\cdot 0.9^4\doteq 0.0046 \\ a=a_0=0.4305 \end{gathered}$$

$b=a_0+a_1+a_2+a_3+a_4=0.4305+0.3826+0.1488+0.0331+0.0046=0.9996$

$c=1-(a_0+a_1+a_2+a_3)=1-(0.4305+0.3826+0.1488+0.0331)=0.005$

$d=a_2=0.1488$

Try another example