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InterviewSolution
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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 : From these days , one day is chosen at random . What is the probability that on that day , the output has (i) no defective part ? (ii) at least 1 defective part ? (iii) not more than 5 defective parts ? (iv) more than 5 , but less than 8 defective parts ? (v) more than 13 defective parts ? |
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Answer» Total number of working days = 200. (i) Let `E_(1)` be the event that the output has 0 defective part on the chosen day . Then , P(event of producing 0 defective part on the chosen day) `= P(E_(1))` `("number of days when the output has 0 defective part")/("total number of working days ")` `= (50)/(200) = (1)/(4) = 0.25` . (ii) Number of days on which the output has at least 1 defective part = 200 - number of days with 0 defective part = 200 - 50 = 150 . Let `E_(2)` be the event that the output has at least 1 defective part on the chosen day . Then , `P(E_(2)) = (150)/(200) = (3)/(4) `. (iii) Let `E_(3)` be the event that the output has not more than 5 defective parts , i.e., 5 or less defective parts , on the chosen day . Then , P(events that the output has not more than 5 defective parts on the chosen day) = P (event that the output has 5 or less defective parts on the chosen day) = `P(E_(3))` `= ("number of days when the output has 5 or less defective parts")/("total number of working days")` `= (50 + 32 + 22+ 18 + 12 + 12)/(200) = (146)/(200) = (73)/(100) = 0.73`. (iv) Let `E_(4)` be the event that the output has more than 5 , but less than 8 defective parts on the chosen day . Then , P (event that the output has more than 5 but less than 8 defective parts on the chosen day) `= P(E_(4))` =`("number of days when the output has 6 or 7 defective parts")/("total number of working days")` `= (10 + 10)/(200) = (20)/(200) = (1)/(10) = 0.1`. (v) Let `E_(5)` be the event that the output has more than 13 defective parts on that day . then , P (event that the output has more than 13 defective parts on that day) `= P(E_(5))` `= ("number of days when the output has more than 13 defective parts")/("total number of working days")` =`(0)/(200) = 0.` |
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