How would I rewrite this using the distributive property?
2 answers:
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Answer:
a)0.976
b)0.00926
c)0.2402
d)0.35
Step-by-step explanation:
Let be an item passed by inspector i
Let Y be the event that there is a fault in an item
The probability that an item has a flaw is 0.1 i.e. P(Y)=0.1
If a flaw is present ,it will be detected by the first inspector with probability 0.92 i.e.
So,
If a flaw is present ,it will be detected by the second inspector with probability 0.7 i.e.
So,
If an item does not have a flaw, it will be passed by the first inspector with probability 0.95 i.e.
So,
If an item does not have a flaw, it will be passed by the second inspector with probability 0.8 i.e.
So,
a)P(found by atleast one inspector | It has flaw )=1-P(found by none inspector | It has flow )
P(found by atleast one inspector | It has flaw )=
P(found by atleast one inspector | It has flaw )=
P(found by atleast one inspector | It has flaw )=0.976
Hence the probability that it will be found by at least one of the two inspectors if it has flaw is 0.976
b)
C)P( two inspectors draw different conclusions on the same item)=
P( two inspectors draw different conclusions on the same item)=0.2402
D)