Question

Question 18
Bulbs are packed in cartons each containing 40 bulbs. Seven hundred cartons were examined for defective bulbs and the results are given in the following table.


$\begin{array}{ccccccccc}\text{Number of}& 0& 1& 2& 3& 4& 5& 6& \text{More}\\ \text{defective}& & & & & & & & \text{than 6}\\ \text{bulbs}& & & & & & & & \\ \text{Frequency}& 400& 180& 48& 41& 18& 8& 3& 2\end{array}$

One carton was selected at random. What is the probability that it has
(i) No defective bulb?

(ii) Defective bulbs from 2 - 6?

(iii) Defective bulbs less than 4?

Answer 1

Total number of cartons, n(S) = 700

(i) Number of cartons which has no defective bulb, $n\left(E_1\right)=400$
$\therefore$ Probability that no defective bulb $=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{400}{700}=\frac{4}{7}$.
Hence, the probability that no defective bulb is $\frac{4}{7}$

(ii) Number of cartons which has defective bulbs from 2 to 6,
$n\left(E_2\right)=48+41+18+8+3=118$
$\therefore$ Probability that the defective bulb from 2 to 6 $=\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{118}{700}=\frac{59}{350}$
Hence, the probability that the defective bulb from 2 to 6 is $\frac{59}{350}$.

(iii) Number of cartons which has defective bulb less than 4,
$n\left(E_3\right)=400+180+48+41=669$.
$\therefore$ The Probability that the defective bulbs less than 4 $=\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{669}{700}$
Hence, the probability that the defective bulb less than 4 is $\frac{669}{700}$.