Question

Question 19
Over the past 200 working days, the number of defective parts produced by a machine is given in the following table.

$\begin{array}{ccccccccccccccc}Numberof& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12& 13\\ defectiveparts& & & & & & & & & & & & & & \\ Days& 50& 32& 22& 18& 12& 12& 10& 10& 10& 10& 8& 6& 6& 2\end{array}$

Determine the probability that tomorrow’s output will have:
(i) No defective part.
(ii) Atleast one defective part.
(iii) Not more than 5 defective parts.
(iv) More than 13 defective parts.

Answer 1

Total number of working days, n(S) = 200

(i) Number of days in which no part is defective, $n\left(E_1\right)=50$
$\therefore$ Probability of no defective part is, $\frac{50}{200}=0.25$

(ii) Number of days in which atleast one part is defective,
$n\left(E_2\right)=32+22+18+12+12+10+10+10+8+6+6+2+2=150$
$\therefore$ Probability that atleast one defective part $=\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{150}{200}=\frac{3}{4}=0.75$
Hence, the probability of least one defective part is 0.75.

(iii) Number of days in which not more than 5 parts aredefective,
$n\left(E_3\right)=50+32+22+18+12+12=146$
$\therefore$ Probability of not more than 5defective parts $=\frac{n\left(E_3\right)}{n\left(S\right)}=\frac{146}{200}=0.73$

(iv) Number of days in which not more than 13defective parts,
$n\left(E_4\right)=0$
$\therefore$ Probability for more than 13 defective parts $=\frac{n\left(E_4\right)}{n\left(S\right)}=\frac{0}{200}=0$
Hence, the probability of more than 13 defective parts is 0.