91Ó°ÊÓ

Combinatorics is a branch of mathematics that deals with counting, arrangement, and combination of objects. In the context of probability, it helps us find how many different ways items can be selected or arranged. For our circuit card problem, combinatorics allows us to determine the possible ways to pick cards for testing.

When calculating combinations, we use a formula that considers how many total items there are and how many you want to choose. The formula is expressed as the binomial coefficient \( \binom{n}{k} \), which stands for "n choose k". It is calculated using:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

Where \( n \) is the total number of items, \( k \) is the number of items to choose, and \( ! \) denotes factorial, meaning the product of all positive integers up to that number.

In our scenario, \( \binom{140}{20} \) calculates how many ways we can choose 20 cards from a lot of 140. Combinatorics is crucial here because it forms the backbone of determining the total possible selections, which is necessary for probability calculations.

Identifying and understanding the role of defective cards is essential in this probability problem. Defective cards are those that do not meet the required standards for functionality. In our scenario, a defective card is one that fails the functional test after being embedded with semiconductor chips.

Knowing the number of defective cards helps us set the parameters for solving the problem. In parts (a) and (b) of the exercise, we encounter different scenarios based on the number of defective cards available:

• In scenario (a), there are 20 defective cards in the lot.
• In scenario (b), only 5 cards are defective.

This information defines the numerator in the probability calculations. By considering how many non-defective cards are left, we can compute the probability that a sample is free of defects. Hence, understanding the defective card count helps in pinpointing the accurate probability for detecting defective items in the selected sample.

Complementary probability simplifies complex probability calculations by focusing on the chances of an event not happening, allowing us to deduce the original probability. The probability of an event occurring is equal to 1 minus the probability of it not occurring. This helps in various scenarios where finding direct probabilities is too complicated.

In our exercise, calculating the probability of selecting at least one defective card would be tedious directly. Instead, we calculate the easier complementary event: selecting no defective cards. The formula used is:

• \( P(\text{at least 1 defective}) = 1 - P(\text{no defective}) \)

For scenario (a), where 20 cards are defective, we find how many ways 20 non-defective cards can be chosen from 120 good ones: \( \binom{120}{20} \). Similarly, for scenario (b), choosing from 135 non-defective cards: \( \binom{135}{20} \).

Thus, by calculating the complementary probability, we can efficiently find the chance of having at least one defective card in the sample, simplifying our task significantly.