Question

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

No of defective parts	0	1	2	3	4	5	6	7	8	9	10	11	12	13
Days	50	32	22	18	12	12	10	10	10	8	6	6	2	2

Determine the probability that tomorrow's output will have not more than 5 defective parts.

• A. $0.33$
• B. $0.5$
• C. $0.66$
• D. $0.73$

Answer 1

Answer: (D)

$0.73$

The no. of possible outcomes $=14$ ....[starting from defect count =0 to defect count =13]

The no. of days required $=50$ to $2$ days.

So, $P\left(\text{no defect}\right)=\left(\frac{50}{200}\right)=0.25$

The probability that there will be at least 1 defect or all 13 defects

$=1-P\left(\text{no defect}\right)=1-0.25=0.75$

The probability that there will be maximum 5 defects $=P\left(0\right)+P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)$

$=\frac{50}{200}+\frac{32}{200}+\frac{22}{200}+\frac{18}{200}+\frac{12}{200}+\frac{12}{200}$

$=\frac{50+32+22+18+12+12}{200}$

$=\frac{146}{200}=\frac{73}{100}=0.73$