1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Please accept the answer if you are satisfied. I am happy to clarify if you have any questions. $\endgroup$ – dlnB Mar 16 at 4:34
  • $\begingroup$ Its very clear . I was almost there but forgot about the mutually exclusive logic $\endgroup$ – prajyal sengupta Mar 16 at 4:36
  • $\begingroup$ 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 $\endgroup$ – prajyal sengupta Mar 16 at 4:49
  • $\begingroup$ 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 $P(A \cup B)=1$. $\endgroup$ – dlnB Mar 16 at 4:52
  • $\begingroup$ cool. Understood. Thanks again $\endgroup$ – prajyal sengupta Mar 16 at 4:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.