Question #144577

A memory chip is produced by two di erent manufacturers. It is known that a chip produced

by manufacturer one is faulty with probability 0:11, while a chip produced by manufacturer

two is faulty with probability 0:17. We also know that manufacturer one supplies 30% of the

chips, while manufacturer two supplies 70% of the chips used by a car company.


(i) What is the probability that given a chip was made by manufacturer two, it is not faulty?

(ii) What is the probability that a chip is not faulty?

What is the probability that given a chip is faulty that it was produced by manufacturer

one?


Expert's answer

P(1st) = 0.3

P(2nd) = 0.7

Then the probability of producing faulty chip by 1st manufacturer is P(f | 1st) = 0.11 and the probability of producing faulty chip by 2nd manufacturer is P(f | 2nd) = 0.17.


i) The probability that chip is not faulty given it was made by 2nd manufacturer :

P(f2nd)=1P(f2nd)=10.17=0.83P(f' | 2nd) = 1-P(f|2nd)=1-0.17=0.83


ii) The probability that a chip is not faulty:

P(f)=1P(f)=1(P(f1st)P(1st)+P(f2nd)P(2nd))=P(f') = 1 – P(f) = 1 – (P(f|1st)\cdot P(1st)+P(f|2nd)\cdot P(2nd))=

=1(0.30.11+0.70.17)=0.848=1-(0.3\cdot 0.11+0.7\cdot 0.17)=0.848


iii) The probability that a chip was produced by 1st manufacturer given it is faulty:

P(1stf)=P(f1st)P(1st)P(f1st)P(1st)+P(f2nd)P(2nd)=0.110.30.110.3+0.170.7=0.217P(1st|f) = \frac{P(f|1st)P(1st)}{P(f|1st)P(1st)+P(f|2nd)P(2nd)}=\frac{0.11\cdot0.3}{0.11\cdot0.3+0.17\cdot0.7}=0.217


Answer:

i) 0.83

ii) 0.848

iii) 0.217


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