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

I believe this question involves the rule of sum and the fundamental counting principle. I think my logic here is wrong, but I hope you could correct it.
(26 P 2 + 26 P 3)(10 P 4) = 8.19*10^7

Given the probability $p=13/20$ that an event will not happen, how do you find the probability that the event will happen?

Janell is having a group of friends over for dinner and is setting the name cards on the table. She has invited 5 of her friends for dinner. How many different seating arrangements are possible for Janell and her friends at the table?

I know that p(a,b) = p(a|b)p(b) = p(b|a)p(a).
But what is the reason for this?
Why don't p(a,b) = p(a|b)p(b|a) since they all involve with each other.

Given the probability $p=0.36$ that an event will happen, how do you find the probability that the event will not happen?

In how many different ways can a pair of dice be rolled to obtain a total of 10?

How many four-digit numbers can be formed from the set $\{0,1,2,3,\dots <!--\; \dots \; -->,10\}$ if zero cannot be the first digit and the given conditions are to be satisfied

1. Repetitions are allowed and the number must be even.
2. Repetitions are allowed and the number must be divisible by 5.
3. The number must be odd and less than 4000 with repetition allowed.

a. my solution is 2*10*10*10= 2000 because 2 is a even number and there are 10 numbers excluding 0 in set {0,1,2,3..10} and it is 4 digits that's why 2*10*10*10 b.same in letter A but 2 is changed into 5 because it must be divisible by 5 so it is 5*10*10*10=5000 c.same to A and B... Only I changed it to 3 so my solution is 3*10*10*10 = 3000

How many three letter arrangements can be formed if a letter is used only once?

I'm trying to apply the counting principle to the following:
"Of 300 people: 35 - bicycle and car. 40 - car and bus. 60 - bicycle and bus. 90 - bicycle. 70 - car. 105 - bus. 25 - bicycle, car, and bus."
I just don't know how this adds up to 300. If i apply the principle, I get:
90 (bicycle) + 105 (bus) + 70 (car) - 35 (bicycle and car) - 40 (car and bus) - 60 (bicycle and bus) + 25 (bicycle, car, and bus).
This equals to 155.
I don't understand how I can get 300 participants from the above, can anyone please help?

How do you calculate permutations using the formula?

A lock has a dial with 50 numbers on it. To open it you must have a 3 digit code. How many possibilities are there if the 3 numbers must be different?

Probability for tossing on heads=0.5 Probability of rolling on odd number on die (1 or 3 or 5)=0.5 As per addition rule (A union B, A or B) that is 0.5+0.5=1 that seems impossible. How could it be? You could very well get a tail and an even-numbered die. Is it a paradox?

How do you find the number of permutations of all the letters of the word SCISSORS which you have no two letters S,S consecutive?

At every basketball game they announce the 5 starting players. How many DIFFERENT ways could they announce them?

I'm trying to prep myself for entrance exams and struggle with the following problem:
Mr. X uses two buses daily on his journey to work. Bus A departs 6.00 and Mr. X always catches that. The journey time follows normal distribution (24,42). Bus B departs at 6.20, 6.30 and 6.40, journey time following normal distribution (20,52), independent of bus A. We assume the transfer from bus A to B won't take time and Mr. X always takes the first bus B available. Calculate probability for Mr. X being at work before 6.55.
So I am aware that the sum of two normal distribution variables follows also normal distribution, but don't know how two proceed with the given departure times of bus B (6.20, 6.30, 6.40) as clearly I can't just calculate the probability of A+B being under 55 minutes.. Any tips how to start with this?

A subway car has 10 individual seats, with 5 in front and 5 in back. From 10 ​passengers, 4 prefers front seat, 3 prefers back seat and the others have no preference. In how many ways can passengers be seated, respecting preferences?
Attempt: I want to solve using only the notion of Fundamental principle of counting:

Let $F$ be each of the 4 passengers who want to sit in the front, $N$ the passengers who have no preference, and $C$ the passengers who want to sit back. Now, since we have 32˙ possibilities to take a passenger without preference (it will be multiplied by the final result) we will take one of the passengers N to sit in the front seat, we will have, by the multiplicative principle:
$5\cdot <!--\; \cdot \; -->4\cdot <!--\; \cdot \; -->3\cdot <!--\; \cdot \; -->2\cdot <!--\; \cdot \; -->1$
Now $N-<!--\; -\; -->1=2$ passengers without preference are left, and they will be exchanged with passengers who want to sit back:
$5\cdot <!--\; \cdot \; -->4\cdot <!--\; \cdot \; -->3\cdot <!--\; \cdot \; -->2\cdot <!--\; \cdot \; -->1$
But if you go on like that, and at the end of it all, multiplying it by 6 I don't get the right result. What am I wrong?
The answer is 43200

How many ways can 3 dictionaries and 5 magazines be arranged on a shelf if there must be a dictionary on each end?

In a game, all points in the (x, y) plane with coordinates obeying $x,y\in <!--\; \in \; -->\mathbb{Z}$ are labelled as belonging to one of three players, i.e., either to Alice, Bob or Carol.
Show that one of the players will possess four points whose vertices form a rectangle.
Here is the problem I've been thinking for days. It seems easy since the coordinates are unlimited. But it is also hard to find pigeonholes I want to divide.
I've encountered several problems like finding a parallelogram on a $n\times <!--\; \times \; -->n$ chessboard with $2n$ pawns. It is relatively simpler to me.

You are told that:
1. $P(P)=0.15$
2. $P(T|P)=0.91$
3. $P(T|\neg P)=0.04$
4. $P$ and $T$ are not independent.

Whats $P(T)$?

I've been really struggling with this one. I've tried substituting a bunch of stuff into the multiplication rule
$P(A\cap <!--\; \cap \; -->B)=P(A)\cdot <!--\; \cdot \; -->P(B|A)$
and the addition rule
$P(A\cup <!--\; \cup \; -->B)=P(A)+P(B)-P(A\cap <!--\; \cap \; -->B).$
I've also tried drawing a conditional probability tree to help me see what's going on, but I just don't seem to be getting anywhere.