A shipment of 20 DVDs has arrived at a video rental store. Based on past experience, the manager knows that 10% of all new DVDs
sent to the store have a visible defect. The manager tells you to begin inspecting the new DVDs one at a time at random until you find the first DVD that has a defect. If 10% of the DVDs have a visible defect in the new shipment, what is the probability that the first DVD that has a defect is the 3rd one that you inspect? (Round your answer to 3 decimal places.)
Here, we want to calculate the probability that the 3rd inspection will be defective.
What this means is that the first two would
not be defective.
Probability of having a defective DVD = 10% = 10/100 = 0.1
Probability of not having a defective DVD = 1 -0.1 = 0.9
So the probability of third being defective = Probability of first not defective * Probability of second not defective * Probability of third defective
This is a very basic area problem. You should try refreshing on some of your multiplication and addition skills. Try reading patterns, maybe. Also learning geometric proofs is VITAL.
