91Ó°ÊÓ

Combinatorics is a field of mathematics focused on counting and arranging objects. In the context of the TV selection problem, we use combinatorics to determine how many ways we can choose a certain number of televisions from the total.

The key formula used here is the combination formula, denoted as \(_{n}C_{k} \). This formula helps us find out how many ways we can select \(k\) objects out of \(n\) without considering the order of selection.
For our exercise, we need \(_{6}C_{2} \) for choosing 2 TVs out of 6, which is calculated as:

\[\binom{6}{2} = \frac{6!}{2!(6-2)!} = 15 \]
The '!' symbol represents factorial, which means product of all positive integers up to that number. So, 6! = 6×5×4×3×2×1.

Using these combinations, we pinpoint exactly how many total ways and how many successful ways are there to pick, giving us the odds or probabilities we need.

The complement rule in probability is a powerful way to understand probabilities of events that are opposite or 'complementary' to each other.
To put it simply, if an event A is what we are interested in, then the complement of A, denoted \( A^c \), is everything that is not A.

For example, in our TV problem, if event A is 'both selected TVs work', then the complement of event A, \( A^c \), is 'at least one TV does not work'. Using this complement relationship helps us easily find probabilities without computing separately.

Using our problem's specifics:

• Probability that both TVs work: \(\frac{2}{5} \)


• Complement event: At least one TV does not work.


• Using the complement rule: \(P(A^c) = 1 - P(A) = 1 - \frac{2}{5} = \frac{3}{5} \)

So, the probability that at least one TV is defective is \( \frac{3}{5} \).

In probability, favorable outcomes are the specific results which satisfy the event we're interested in. Determining these outcomes takes us a step closer to finding the overall probability.

In the TV selection example, we focused on the event that both chosen TVs are working.

• Out of six TVs, we know four are working.


• Using combinatorics, we calculated \( \binom{4}{2} = 6 \) as the number of ways to choose 2 working TVs from 4. These are our favorable outcomes for the event 'both TVs work'.


• For the total possible outcomes, choosing any 2 TVs from 6 was \( 15 \).


• The probability formula becomes: \(\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{15} = \frac{2}{5} \)

Here, favorable outcomes focused on the probability of getting both working TVs out of our random selection, helping clarify different scenario possibilities.