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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 more than 13 defective parts
बेरीज
उत्तर
Total number of working days, n(S) = 200
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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