A manufacturer produces ultra cheep light bulbs, in lots of 75, in which 12% of the bulbs are defective. An inspector selects a box and randomly selects 7 bulbs. What is the probability that no more than 2 of the 7 bulbs are defective?
my wild guess is 29/175 :P let me know if i am right i going to bed
wrong :( i know the answer : 0.9664545 I just dont at all know how to go about finding it.
aaa i have to give it another look at another time. I going to work :P
Probability of no more than 2 defects Probability of a defective bulb = .12 Probability of good buld = .88 Binomial Probabilty No more than 2 defects means 0 defects, 1 defect, or 2 defects Zero Defects: C(7,0) (.12)^0 * (.88)^7 One Defect: C (7.1) (.12)^1 * (.88)^6 Two Defects: C (7,2) (.12)^2 * (.88)^5 Probability of No More than Two Defects = C(7,0) (.12)^0 * (.88)^7 + C (7.1) (.12)^1 * (.88)^6 + c (7,2) (.12)^2 * (.88)^5 =
= 0.9583611609088
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