High school probability Uncovered: Discover the Secrets with Plainmath's Expert Help

You have 20 different neckties in your wardrobe. How many combinations of three ties could you choose?

How many different ways are there to arrange the 6 letters in the word SUNDAY?

In my textbook MP is strictly reserved to counting lists. Does what I do below to count subsets work?
Consider 3 men(Ace, Bob, Corry) and 3 women(Ann, Beth, Candace). Suppose we need to choose a team with 2 men and 2 women in it. An example team is $\{\text{Ace, Ann, Beth, Bob}\}$ which is the union of $\{\text{Ace, Bob}\}\{\text{Beth, Ann}\}.$ So we simply count 2−lists whose first element is a set of two men and whose second element is a set of two women. We get the same number of 2−lists if their first element is a set of two women. There are $(\genfrac{}{}{0ex}{}{3}{2})=8$ two-men subsets and that many two-women subsets. So there are 8 choices for the first element and 8 choices for the second one. In all there are $8^2={(\genfrac{}{}{0ex}{}{3}{2})}^2=64$ two element lists = teams with two men and two women in each.

Three cards are drawn from a standard deck of 52 cards. What is the possibility that a jack, a queen, and a king are selected, drawn in succession without?

Students are chosen in groups of 6 to tour a local business. How many ways can 6 students be selected from 3 classes totaling 53 students?

This is one of our probability exercise:

We have an urn with m green balls and n yellow balls. Two balls are drawn at random. What is the probability that the two balls have the same color?
(a) Assume that the balls are sampled without replacement.
(b) Assume that the balls are sampled with replacement.
(c) When is the answer to part (a) larger than the answer to part (b)? Justify your answer. Can you give an intuitive explanation for what the calculation tells you?

For (a), I think there are only two possible outcomes of this experiment, which are either two green balls or two yellow balls. Since they are independent on each other, so we can use the addition rule. Since it is without replacement, so we can directly use combinations.
$P(\text{same color})=\frac{(\genfrac{}{}{0ex}{}{m}{1})\times <!--\; \times \; -->(\genfrac{}{}{0ex}{}{m-<!--\; -\; -->1}{1})+(\genfrac{}{}{0ex}{}{n}{1})\times <!--\; \times \; -->(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->1}{1})}{(\genfrac{}{}{0ex}{}{m+n}{2})}$
The difference between replacement and no replacement is how the candidate pool changes after each action. With replacement, we don't need to modify the number of candidates anymore.
$P(\text{same color})=\frac{(\genfrac{}{}{0ex}{}{m}{1})\times <!--\; \times \; -->(\genfrac{}{}{0ex}{}{m}{1})+(\genfrac{}{}{0ex}{}{n}{1})\times <!--\; \times \; -->(\genfrac{}{}{0ex}{}{n}{1})}{(m+n{)}^2}$
However, I have no clues how to solve (c). I tried to use algebra to simplify my answer for (a) and (b), but it doesn't help.
Any hints or suggestions would be appreciated!

A necklace is to be made from 3 red, 2 black and 2 white beads. How many possible configurations are there?

There are 9 different colors of paint to choose from. Out of the 9 colors, how many ways can 4 different colors be chosen?

How do you simplify 7P5?

How many different selections of 4 books can be made from a bookcase displaying 12 books?

You have eight different suits to choose from to take on a trip. How many combinations of three suits could you take?

In how many ways can we list out a string of seven numbers (using only the numbers 1, 2, and 3) such that they sum to 10?

Machine 1 is currently working. Machine 2 will be put in use at time $t$ from now. If the lifetime of machine $i$ is exponential with rate ${\mathrm{\lambda <!--\; \lambda \; -->}}_i,i=1,2,$ what is the probability that machine 1 is the first machine to fail?

The solution is as follows:
$\mathbb{P}(M_1<<!--\; <\; -->M_2)=\mathbb{P}(M_1<<!--\; <\; -->M_2|M_1<t)\cdot <!--\; \cdot \; -->\mathbb{P}(M_1<t)+\mathbb{P}(M_1<M_2|M_1>t)\cdot <!--\; \cdot \; -->\mathbb{P}(M_1>t).$
Now, I understand this solution. It states, in light of Bayes's Rule*, that we desire the probability that the lifetime of machine 1 is less than the lifetime of machine 2 and the lifetime of machine 1 is less than $t$ in addition to the probability that the lifetime of machine 1 is less than the lifetime of machine 2 and the lifetime of machine 1 is greater than $t$. That all makes sense to me. My question, however, is why is it not simply
$\mathbb{P}(M_1<<!--\; <\; -->M_2)=\mathbb{P}(M_1<t)+\mathbb{P}(M_1<M_2|M_1>t).$
Linguistically, this seems to satisfy the solution to the problem; i.e., "the probability that the lifetime of machine 1 is less than t, in addition to the probability that the lifetime of machine 1 is less than that of machine 2 given that the lifetime of machine 1 is greater than t. I understand that the first term here is actually the same as the solution's, and that the solution formulates it as it does for illustrative purposes; but the second term certainly has a different value than that of the solution's.

I hope my question here makes sense. If any further clarification is needed, please let me know. Also, I understand that this exact problem appears elsewhere on stackexchange, but it was more about the computation.

*The Bayes's Rule to which I'm referring is, of course, the following formulation:
$\mathbb{P}(AB)=\mathbb{P}(A|B)\cdot <!--\; \cdot \; -->\mathbb{P}(B)$

In how many ways can 6 people be arranged in a line?

How many different permutations can you make with the letters in the word seventeen?

What is 6P3?

Given the probability $p=1/3$ that an event will happen, how do you find the probability that the event will not happen?

There are 40 numbers in the Louisiana Lotto game. In how many ways can a player select six of the numbers?

Show that ${\displaystyle (n+2)!+(n+1)!+n!}$?

How do you calculate permutations of numbers?