Finding the probability for non defective battery A car manufacturer purchases car batteries from two different suppliers A and B.
Suppose supplier A provides 60% of the batteries and supplier B provides the rest. If 6%
of all batteries from supplier A are defective and 4% of the batteries from supplier B are
defective. Determine the probability that a randomly selected battery is not defective.
Solution:
P(Selecting not a defective battery) =1- P(Selecting a defective battery) or P(Selecting a defective battery given A is defective) or P(selecting a defective battery given B is defective)
P(Selecting a defective battery) = (.6*.06)+(.4*.04) = 0.052
P(Selecting a defective battery given A is defective) = (.6*.06) = 0.036
p(Selecting a defective battery given B is defective) - (.4*.04) = 0.016
Therefore
P(Selective a not defective battery is ) = 1 - (0.052+0.036+0.016) = 0.9284
Please let me know if this is correct. Please provide you comment and advice on my solution . I am a bit confused whether it is correct or not.
 A: Your solution is incorrect. This problem requires a simple application of Bayes' rule.
$$P(\text{selecting a defective battery})=P(\text{(battery comes from A and battery is defective) or (battery comes from B and battery is defective)}).$$
Since (battery comes from A and battery is defective) and (battery comes from B and battery is defective) are mutually exclusive events, and more specifically form a partition of the set of all batteries, it follows
$$P(\text{selecting a defective battery})=P(\text{Battery comes from A and is defective}) + P(\text{battery comes from B and battery is defective}).$$
Bayes' rule tells us
$$P(\text{battery comes from A and battery is defective})=P(\text{battery is defective} | \text{battery comes from A})P(\text{battery comes from A}) = 0.6*0.06=.036.$$
and 
$$P(\text{battery comes from B and battery is defective})=P(\text{battery is defective} | \text{battery comes from B})P(\text{battery comes from B})=0.4*0.04=0.016.$$
Therefore
$$P(\text{selecting a defective battery})=0.036+0.016=0.054.$$
This gives
$$P(\text{selecting a non-defective battery})=1-0.054=0.946.$$
