the problem asks what is the probability of accepting and rejecting a shipment, for accepting i got 0.4912. would the probability of rejecting be 1-0.4912? thank you.
if you wanted to know the entire problem it is:
suppose that you have just received a shipment of 20 modems. although you dont know this, 3 of the modems are defective. to determine wheather you will accept the shipment, you randomly select 4 modems and test them. if all 4 modems work, you accept the shipment. otherwise the shipment is rejected. what is the probability of accepting the shipment? what is the probability of rejecting the shipment?
if you wanted to know the entire problem it is:
suppose that you have just received a shipment of 20 modems. although you dont know this, 3 of the modems are defective. to determine wheather you will accept the shipment, you randomly select 4 modems and test them. if all 4 modems work, you accept the shipment. otherwise the shipment is rejected. what is the probability of accepting the shipment? what is the probability of rejecting the shipment?
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Probability of rejecting the shipment = 1 - prob of accepting the shpment. Because there are only 2 exhaustive cases, accepting and rejecting.
Now, total ways of selecting 4 modems= 20c4.
There are 17 good ones, and 3 defectives.
Ways of selecting good ones = 17c4.
So probabilty of accepting = 17c4/20c4 = .491
so prbablty of rejecting = .509
Now, total ways of selecting 4 modems= 20c4.
There are 17 good ones, and 3 defectives.
Ways of selecting good ones = 17c4.
So probabilty of accepting = 17c4/20c4 = .491
so prbablty of rejecting = .509
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there are 17 modems that work
17c4 / 20c4 = 2380 / 4845
0.49122807017543859649122807017544
49.12%
yes, it would be 50.88%
17c4 / 20c4 = 2380 / 4845
0.49122807017543859649122807017544
49.12%
yes, it would be 50.88%