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Question
Over the past 200 working days, the number of defective parts produced by a machine is given in the following table:
| Number 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
- no defective part
- atleast one defective part
- not more than 5 defective parts
- more than 13 defective parts
Solution
Total number of working days, n(S) = 200
i. Number of days in which no defective part is,
n(E1) = 50
Probability that no defective part = `(n(E_1))/(n(S))`
= `50/200`
= `1/4`
= 0.25
ii. Number of days in which atleast one defective part is,
n(E2) = 32 + 22 + 18 + 12 + 12 + 10 + 10 + 10 + 8 + 6 + 6 + 2 + 2 = 150
∴ Probability that atleast one defective part = `(n(E_2))/(n(S))`
= `150/200`
= `3/4`
= 0.75
iii. Number of days in which not more than 5 defective parts,
n(E3) = 50 + 32 + 22 + 18 + 12 + 12 = 146
∴ Probability that not more than 5 defective parts
= `(n(E_3))/(n(S))`
= `146/200`
= 0.73
iv. Number of days in which more than 13 defective parts,
n(E4) = 0
= `(n(E_4))/(n(S))`
= `0/200`
= 0
Hence, the probability that more than 13 defective parts is 0.
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Over the past 200 working days, the number of defective parts produced by a machine is given in the following table:
| Number 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
