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प्रश्न
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 atleast one defective part
योग
उत्तर
Total number of working days, n(S) = 200
Number of days in which atleast one defective part is
n(E2) = 32 + 2 + 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
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