# 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.

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 $$P(\text{selecting a non-defective battery})=1-0.054=0.946.$$

• Thanks got it.. – prajyal sengupta Mar 16 at 4:34
• Please accept the answer if you are satisfied. I am happy to clarify if you have any questions. – dlnB Mar 16 at 4:34
• Its very clear . I was almost there but forgot about the mutually exclusive logic – prajyal sengupta Mar 16 at 4:36
• The only question I had was it is a mutually exclusive event because both A and B can supply battery to the manufacture at the same time. What I mean is 60% of the battery is coming from A and 40% coming from B. Let me know your thoughts – prajyal sengupta Mar 16 at 4:49
• I edited my answer to be more clear. Writing $P(D)=P(A \cap D)+P(B \cap D)$ actually requires more than just mutual exclusivity. It requires $A$ and $B$ to make up a partition of the entire set of batteries, i.e. $P(A \cap B)=0$ and $P(A \cup B)=1$. – dlnB Mar 16 at 4:52