91Ó°ÊÓ

Combinatorial analysis is a fundamental aspect of probability theory in statistics. It involves counting, arranging, and selecting elements within a finite set in specific ways.
In this exercise, we use combinatorial analysis to calculate the number of ways to select 5 televisions from a shipment of 100. The formula we use is the binomial coefficient, noted as \(\binom{n}{k}\). It calculates the number of ways to choose \(k\) items from \(n\) items without regard to the order. More formally, it's given by:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]In our case,

• \(n = 100\)
• \(k = 5\)

Substituting, we find: \[ \binom{100}{5} = \frac{100!}{5!(100-5)!} \]This gives us the total number of ways to choose 5 televisions out of 100. Similarly, for choosing 5 non-defective televisions from 94 non-defective ones:
\[ \binom{94}{5} = \frac{94!}{5!(94-5)!} \]This groundwork simplifies finding the probability of an event by providing clear values to compare.

In probability and statistics, sampling without replacement refers to drawing elements from a set such that they are not returned to the set after being chosen.
In our example, when we select a television from the shipment, if it is defective or not, it is not returned to the shipment before the next selection. This impacts the calculation of probabilities because each draw changes the composition of the set.
For instance, the total number of ways to choose 5 televisions out of 100 changes if we were to replace the televisions back into the set after drawing. Thus, without replacement, each selection affects the subsequent ones, hence changing the probabilities continually. Here's how it fits in the given problem: If the first selected TV is non-defective, we have 4 more selections to make from the remaining 99 TVs.
This approach ensures that our probability calculations account for the decreasing number of TVs after each draw.

Determining the probability of defective items involves understanding the ratio of favorable outcomes to the total possible outcomes.
In our case, we're interested in the probability that all 5 selected televisions are non-defective. Initially, we know 94 out of the 100 televisions are non-defective. The favorable outcome is selecting 5 non-defective ones. The probability formula we use here is:
\[ P(\text{all 5 televisions work}) = \frac{\binom{94}{5}}{\binom{100}{5}} \]
This ratio gives us the likelihood of this specific scenario happening—the numerator represents the favorable combinations (5 non-defective TVs out of 94) while the denominator represents the total combinations (5 TVs out of 100).
Numerically:
\[ \binom{100}{5} \text{values roughly as} \ 75,287,520 \]
\[ \binom{94}{5} \text{values roughly as} \ 71,658,950 \]
Thus, the probability:
\[ P(\text{all 5 televisions work}) = \frac{71,658,950}{75,287,520} \text{approximates to} \ 0.951 \]
This calculated probability indicates there is a 95.1% chance that all 5 randomly selected televisions from the shipment will be non-defective, leading us to accept the shipment.