Question

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.

Number of defective bulbs	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$Morethan6$
Frequency	$400$	$180$	$48$	$41$	$18$	$8$	$2$	$2$

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

Step 1: Find the probability of bulb drawn at a random from carton has no defective bulbs:

Number of defective bulbs	Frequency
$0$	$400$
$1$	$180$
$2$	$48$
$3$	$41$
$4$	$18$
$5$	$8$
$6$	$2$
$Morethan6$	$2$
Total	$699$

Let $E$ be the event of drawing bulbs from the carton having no defective pieces.

Number of cartons with zero defective bulbs are $n\left(E\right)=400$.

Total number of observations taken into account $n\left(S\right)=699$

We know that, $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

By substituting the values we get, $P\left(E\right)=\frac{400}{699}$.

Step 2: Find the probability of bulb drawn at a random from carton has $2-6$ defective bulbs:

Let $E_1$ be the event of drawing bulbs from the carton having $2-6$ defective pieces.

Number of cartons with $2-6$ defective bulbs are $n\left(E_1\right)=48+41+18+8+2=117$

Total number of observations taken into account $n\left(S\right)=699$

We know that, $P\left(E_1\right)=\frac{n\left(E_1\right)}{n\left(S\right)}$

By substituting the values we get, $P\left(E_1\right)=\frac{117}{699}=\frac{39}{233}$.

Step 3: Find the probability of bulbs drawn at a random from carton has less than $4$ defective bulbs:

Let $E_2$ be the event of drawing bulbs from the carton containing less than $4$ defective pieces.

Number of cartons with less than $4$ defective bulbs are: $n\left(E_2\right)=400+180+48+41=669$

Total number of observations taken into account $n\left(S\right)=699$

We know that, $P\left(E_2\right)=\frac{n\left(E_2\right)}{n\left(S\right)}$

By substituting the values we get, $P\left(E_2\right)=\frac{669}{699}=\frac{223}{233}$

Hence, the probability of drawing carton with no defective bulbs, having $2-6$ defective bulbs and less than $4$ defective bulbs are $\frac{400}{699},\frac{39}{233}and\frac{223}{233}$ respectively.