A chain of video stores sells three different brands of DVD players. Of its DVD player sales, 50% are brand 1 (the least expensive), 30% are brand 2, and 20% are brand 3. Each manufacturer offers a 1-year warranty on parts and labor. It is known that 25% of brand 1's DVD players require warranty repair work, whereas the corresponding percentages for brands 2 and 3 are 20% and 10%, respectively. I. What is the probability that a randomly selected purchaser has bought a brand 1 DVD player that will need repair while under warranty? Ii. What is the probability that a randomly selected purchaser has a DVD player that will need repair while under warranty? Iii. If a customer returns to the store with a DVD player that needs warranty repair work, what is the probability that it is a brand 1 DVD player? A brand 2 DVD player? A brand 3 DVD player?

I. To find the probability that a randomly selected purchaser has bought a brand 1 DVD player that will need repair while under warranty, we can multiply the probability of a customer buying a brand 1 DVD player by the probability that a brand 1 DVD player will need repair.

The probability of buying a brand 1 DVD player is 50% or 0.50, and the probability that a brand 1 requires warranty repair is 25% or 0.25.

So the probability is 0.50 * 0.25 = 0.125 or 12.5%.

II. To calculate the probability that a randomly selected purchaser has a DVD player that will need repair while under warranty, we need to consider all the brands and their respective probabilities of requiring repairs. We can use the law of total probability.

For brand 1, we have already calculated this probability as 0.125.

For brand 2, with 30% sales and a 20% repair rate, the probability is 0.30 * 0.20 = 0.06 or 6%.

For brand 3, with 20% sales and a 10% repair rate, the probability is 0.20 * 0.10 = 0.02 or 2%.

Now, we need to add up these probabilities to find the total probability of a DVD player needing repair:

Total probability = P(Brand 1 needs repair) + P(Brand 2 needs repair) + P(Brand 3 needs repair) = 0.125 + 0.06 + 0.02 = 0.205 or 20.5%.

III. To find the probability that a returning DVD player needing repairs is of a specific brand, we use Bayes' theorem. Here, we need to calculate the updated probability based on the fact that we know the DVD player needs repairs.

The probability that a returning DVD player is a brand 1, given that it needs repair, is:

P(Brand 1 | Needs repair) = (P(Needs repair | Brand 1) * P(Brand 1)) / P(Needs repair) We already have all these values: P(Needs repair | Brand 1) is 0.25, P(Brand 1) is 0.50, and P(Needs repair) is 0.205.

So P(Brand 1 | Needs repair) = (0.25 * 0.50) / 0.205 ≈ 0.6098 or 60.98%.

Similarly, we can find this probability for brand 2 and brand 3:

P(Brand 2 | Needs repair) = (P(Needs repair | Brand 2) * P(Brand 2)) / P(Needs repair) = (0.20 * 0.30) / 0.205 ≈ 0.2927 or 29.27%.

P(Brand 3 | Needs repair) = (P(Needs repair | Brand 3) * P(Brand 3)) / P(Needs repair) = (0.10 * 0.20) / 0.205 ≈ 0.0976 or 9.76%.

Extra: The problems we are solving here use concepts of probability including the basic probability formula, the law of total probability, and Bayes' theorem. These concepts help us understand how likely certain events are to happen based on given conditions.

1. Basic probability is simply the chance that a certain event will occur, and it is found by dividing the number of favorable outcomes by the total number of possible outcomes.

2. The law of total probability allows us to find the probability of an event when we can break down that event into distinct, non-overlapping scenarios.

3. Bayes' theorem is a formula that describes how to update the probabilities of hypotheses when given evidence. It's particularly used to find the probability of a cause given that an effect is observed (the reverse of the usual conditional probability).

These concepts are widely used in various fields, not just in retail analysis, but also in areas like health studies, risk assessment, and much more. They provide a mathematical means to make informed decisions based on data.