3
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 more than 13 defective parts.
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 |
Open in App
Solution
The correct option is A 0
The no. of possible outcomes =14 ...[starting from defect count =0 to defect count =13]
The no. of days required =50 to 2 days.
As the maximum defects possible would be 13, hence the event of more than 13 defects could be treated as an impossible event.
So, the probability of more than 13 defects =0.
The no. of possible outcomes =14 ...[starting from defect count =0 to defect count =13]
The no. of days required =50 to 2 days.
As the maximum defects possible would be 13, hence the event of more than 13 defects could be treated as an impossible event.
So, the probability of more than 13 defects =0.
Suggest Corrections
1
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
