Probability question. will choose BEST answer...need help :(?

w need to find out the size of the lot to compute the defectives. then we subtract defectives from the the lot size and divide by the lot size , multiply by 100 to obtain the %of a good cartridges per lot.

y= lot size

7 samples per lot are taken so what is the size of the lot?

y / 7 -1 = 1

y = 14

the % defective is 1.7% , 14(1.7)/100 = 0.238 defects per lot

14 - 0.238 = 13.76 good cartridges, ----> 13.76/14*100 = 98.3% good cartridges per lot which is the

probability of success in any one trial. So the probabilities the shipment will be accepted are unknown.But there is more on the calc.of the probability of acceptance of the shipment.For we have to make use of a formula called the Binomial Probability Formula which give the probability of sucess given the values already calculated.

P(x) = nCx *p^x *q^(n-x) where the definitions are:

n = number of trials = 7

x= number of succees among trials = 6

p = probability of success in any one trial = 98.3%

q= probability of failure in any one trial ( 1-p )

nCx = combinations of n items , choose x

from here is a matter of substitution: nCx = 7C6 = 7! / 6 !(7-6)! = 7*6! / 6! (1)! = 7!

so we have, P(6) = 7 *(0.983)^6 * (0.017)^(7-6) = 6*(0.9022)(0.017) = 0.107 = 10.7% which is the

the probability of the shipment being rejected, then 100- 10.7

= 89.3 % being accepted.

To be accepted, all 7 cartridges in the sample must not be defective,

and the probability is (1 - 0.017)^7 = 0.8869.

P[defective] = 0.017

so P[ok] = 1- 0.017 = 0.983

P[ all 7 ok = shipment accepted ]

= 0.983^7

= 0.8869 or 88.69%

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