Question : Over the past 200 working days, the number of defective parts produced by a machine is given in the following table : Determine the probability that tomorrow's output will have :
| 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 |
(i) No defective part
(ii) Atleast 1 defective part
(iii) More than 13 defective parts
Doubt by Aditi
Solution :
Total number of working days = 200
(i) No. of days in which there is no defective part produced by the machine = 50
P(Event) = No. of favourable outcomes)/(Total No. of possible outcomes)
P(No defective part)
= 50/200
= 1/4
= 0.25
(ii) No. of days in which the machine has produced atleast 1 defective part
= 50+32+22+18+12+12+10+10+10+8+6+6+2+2
= 200-50
= 150
P(atleast 1 defective part)
= 150/200
= 3/4
= 0.75
(iii) No. of days in which the machine has produced more than 13 defective parts = 0
P(More than 13 defective parts)
= 0/200
= 0