Addition Rule of Probability: Example Problems

I am trying to improve my stat skills from a book that gives an exercise, I cannot get my head around.<br.It goes like this:<br.There are 5 green balls and 1 red ball in a box. We draw four times randomly with replacement. What is the probability that we draw at least two red balls?<br.My guess would be<br.(1/6)^2 + (1/6)^3 + (1/6)^4 because the probability of at least 2 red balls out of 4 draws must be equal to drawing two or three or four times a red ball. Hence I would use the addition rule, which gives 0.03 = 3 percent<br.However, the book says it is 13 percent, and gives the following explanation (150+20+1)/1296.<br.Why is this so?<br.My attempt to trace back the terms:<br.I can possibly see how it got the first and third term in the fraction: The first could be (5/6)^2 = 25/36, which is the probability of drawing two green balls; and the third could be (1/6)^4 = 1/1296, which is the probability of drawing four red balls. I have no idea though where they could possibly get the 20 from.

We have a condensed deck of 40 cards containing only the denominations from Ace through 10. We draw four cards uniformly at random (without replacement).
1. What is the probability that we draw four distinct denominations, such as one Ace, one five, one six and one seven.
2. What is the probability that we draw at least three hearts.
For 1, let's say we divide the the 40 cards into 10 piles of 4 cards each (of each kind Ace to 10). Then there are 10C4 ways to choose 4 piles and 4C1 ways to choose a card from each pile, so the probability should be ((10C4)(4)(4C1))/(40C4) = 0.036. Is this correct or am I missing something?
For 2, I think something like 1 - Pr(choosing exactly 2 hearts) could work, but I don't know where to go from here.

I know the independent concept and multiplication rules.

Example Flip Coin

5 times flip, how many times success of Head?
Ans : $\frac{1}{2}\cdot <!--\; \cdot \; -->\frac{1}{2}\cdot <!--\; \cdot \; -->\frac{1}{2}\cdot <!--\; \cdot \; -->\frac{1}{2}\cdot <!--\; \cdot \; -->\frac{1}{2}=\frac{1}{32}$

Why do we use multiplication rules, not other rules?
I know if we use addition then probability 2.5 which is invalid because probability should be between 0 to 1. Same for Subtraction and division.
But, I am still confused why we use multiplication? What is the basic and core logic behind this?
Can anyone tell me the concept of this?

If I toss a coin 3 times and want to know the probability of at least one head, I have understood that the answer is $1-<!--\; -\; -->{0.5}^3=99\mathrm{\%<!--\; \%\; -->}$. However, why cannot I not use the additon rule $P(A\cup <!--\; \cup \; -->B)=P(A)+P(B)-<!--\; -\; -->P(A\cap <!--\; \cap \; -->B)$, i.e. $0.5+0.5+0.5-<!--\; -\; -->{0.5}^3$?

The general addition rule for non-mutually exclusive is
P(A OR B) = P(A) + P(B) - P(A AND B);
Now I know how to calculate P(A) and P(B), but how do you calculate P(A AND B)?
Example: What is the probability of rolling a 5 or odd number? To find P(A AND B) in this case is very simple, we know without calculations that it is 1, right? But how to find P(A AND B) when there is a lot of counting to do... like lets say there are 9999 slots which are numbered, some are black some are white, some are even some are odd. Now to find the probability of getting a black and even when a die is rolled upon the slots, how to find P(A AND B)? or we just have to count it just like that?

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!

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:
$\mathbb{P}(AB)=\mathbb{P}(A|B)\cdot <!--\; \cdot \; -->\mathbb{P}(B)$

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.

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?

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?

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.

If you roll 5 dices and get three of the same dices, is it better to leave one and roll one or roll two dices?
My friend is saying that it is better to leave one and roll another, since that will give you a full house in 1/6 probability.
However, I thought it is better to roll both different dices since that will give 5/36 full house and 1/36 yahtzee.
Am I wrong in thinking that it is always better to roll two dices since yahtzee scores more than full house given that both full house and yahtzee are open?

Came across this,
$\mathrm{\varphi <!--\; \varphi \; -->}=$ probability of event
$p(y,\mathrm{\varphi <!--\; \varphi \; -->})={\mathrm{\varphi <!--\; \varphi \; -->}}^y(1-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->}{)}^{(1-<!--\; -\; -->y)}=Exp(\log <!--\; \; -->({\mathrm{\varphi <!--\; \varphi \; -->}}^y(1-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->}{)}^{(1-<!--\; -\; -->y)}))$ and now, somehow $Exp(\log <!--\; \; -->({\mathrm{\varphi <!--\; \varphi \; -->}}^y(1-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->}{)}^{(1-<!--\; -\; -->y)}))=Exp[\log <!--\; \; -->(\frac{\mathrm{\varphi <!--\; \varphi \; -->}}{1-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->}})y+\log <!--\; \; -->(1-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->})]$. I looked to use the power rule for logs to bring down the $y(1-<!--\; -\; -->y)$ in front, the product rule for logs to get an addition to the equation, and the subtraction rule for logs to get the division part of the expression. Upon simplifying, I don't yield the desired expression

My attempt is shown below:
$\begin{array}{rl}Exp(\log <!--\; \; -->({\mathrm{\varphi <!--\; \varphi \; -->}}^y(1-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->}{)}^{(1-<!--\; -\; -->y)}))& =Exp[(1-<!--\; -\; -->y)\log <!--\; \; -->({\mathrm{\varphi <!--\; \varphi \; -->}}^y(1-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->}))]\\ & =Exp[(1-<!--\; -\; -->y)\log <!--\; \; -->({\mathrm{\varphi <!--\; \varphi \; -->}}^y)+\log <!--\; \; -->(1-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->})]\\ & =Exp[(1-<!--\; -\; -->y)y\log <!--\; \; -->(\mathrm{\varphi <!--\; \varphi \; -->})+\log <!--\; \; -->(1-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->})]\end{array}$
So, now I'm not sure if this can arrive at the right answer

The question is from a standardized test. It is based around a type of probability. I have not taken a class that involves probability, as it is not taught in our standard curriculum. I am not positive on if my partial efforts are correct, but I and many others I know would appreciate some clarification. The problem goes (keep in mind this is from memory, I do not have answer choices memorized. So if you are unsure I cannot provide any choices to go from.)

"You have a jar of 10 cookies. In this jar are 5 chocolate chip cookies, 3 oatmeal cookies, and 2 raisin cookies. You reach in at random and take out a cookie. Without replacing the first cookie you pulled, you pull out a second. What is the probability that the first and second cookies will be the same type?"

As stated, this is from memory. I have spoken with friends to confirm and all agree this is the exact problem. We may be wrong, so if this is mathematically impossible to solve, be sure to let me (us) know.

From my very limited knowledge of probability, I cannot see a way to form one probability from three. You would take the probability of the first pick (for each type of cookie), 5/10, 3/10, and 2/10, and multiply them respectively by 4/9, 2/9, and 1/9. That would give you the probability of picking the same for each type.

How would you form one single probability from those three (if possible)?

As I said. I could be wrong, as if this is just not possible, then I obviously am not recalling it clearly enough. Thanks in advance.

This example is from the book first course in probability Example 4a.
Let $X$ denote a random variable that takes on any of the values −1, 0, and 1 with respective probabilities $P\{X=-1\}=.2P\{X=0\}=.5>P\{X=1\}=.3$.
Solution.Let $Y=X^2$. Then the probability mass function of $Y$ is given by
$P\{Y=1\}=P\{X=-1\}+P\{X=1\}=.5$
$P\{Y=0\}=P\{X=0\}=.5$
What rule is used to compute $P\{Y=1\}$? When $Y=X^2$ Why addition is used ?

Having been reprimanded for posting a question on the wrong site, I hope I'm not transgressing this time. In the addition rule for non mutually exclusive events $P(A\cup <!--\; \cup \; -->B)=P(A)+P(B)-<!--\; -\; -->P(A\cap <!--\; \cap \; -->B)$, while $P(A\cap <!--\; \cap \; -->B)$ can be determined by observation, why cannot it always be calculated from $P(A)\cdot <!--\; \cdot \; -->P(B)$?

Example: A lottery box contains 50 lottery tickets numbered 1 to 50. If a lottery ticket is drawn at random, what is the probability that the number drawn is a multiple of 3 or 5? $P(X\cup Y)=P(X)+P(Y)-<!--\; -\; -->P(X\cap Y)$ Therefore,
$P(XUY)=8/25+1/5-<!--\; -\; -->3/50=(16+10-<!--\; -\; -->3)/50$
$=23/50$
But using $P(X\cap Y)=P(X)\cdot <!--\; \cdot \; -->P(Y)=8/25\cdot <!--\; \cdot \; -->1/5=8/(25\cdot <!--\; \cdot \; -->5)=8/125$ which is NOT 3/50?
There are situations where $P(X\cap Y)=P(X)xP(Y)$ works perfectly, but not in others.

In Texas holdem, one is dealt a Decent Hand (Any pocket pair or any two broadway cards) $\sim <!--\; \sim \; -->$15 percent of the time. If there are three people left in the hand, I can use the probability addition rule to say at least one of those three people left will show up with a Decent Hand $\sim <!--\; \sim \; -->$45 percent of the time, correct? What about when there are 10 people left? Does someone show up with a Decent Hand $\sim <!--\; \sim \; -->$150% of the time?
Edit: Why does the addition rule not suited for this case?

20 blue balls and 11 yellow, drawing 6 times with no replacement, what is the chance that at least one is yellow OR the first two draws are the same?

I'm not sure my intuition for the solution of the problem is correct, in particular, I have some issue with the addition rule part.I started out this problem by thinking about the two scenarios:

1) the first two are yellow
2) the first two are blue

I'm a bit confused about the first scenario: P(at least 1 yellow) OR P(first two are the same) if the first two are yellow, that also satisfy the P(at least 1 yellow) so in this case does the probability equal to 1?
Is this similar to flipping a coin P(head) or P(tail)? Or perhaps it's only a "double counted intersection" that we have to subtract out?
I also thought about using compliment to solve this problem but I'm not quite sure if it works this way:
1- [P(no yellow) or P(first two are different)]

What is $P(A\cup <!--\; \cup \; -->(B\cap <!--\; \cap \; -->C))$?

The question says it all. I know
$P(A\cap <!--\; \cap \; -->(B\cup <!--\; \cup \; -->C))=P(A\cap <!--\; \cap \; -->B)+P(A\cap <!--\; \cap \; -->C).$
Would this mean,
$P(A\cup <!--\; \cup \; -->(B\cap <!--\; \cap \; -->C))=P(A\cup <!--\; \cup \; -->B)+P(A\cup <!--\; \cup \; -->C)?$
Just want to make sure.

Alfonso and Colin each bought one raffle ticket at the state fair. If 50 tickets were randomly sold, what is the probability that Alfonso got ticket 14 and Colin got ticket 23?

The answer should be $\frac{1}{2450}$ which presumably comes from $\frac{1}{50}\times <!--\; \times \; -->\frac{1}{49}$. But it seems that the order does not count. I did not assume that Alfonso got ticket 14 first then Colin got ticket 23 second.

Update: What is wrong with this reasoning. When I said that I did not assume order, I meant that it's possible

1. Alfonso got ticket 14 first, then Colin got ticket 23,
2. Colin got ticket 23 first, then Alfonso got ticket 14.

Both of these possibilities are possible before the tickets are given out, so we can make an 'or' statement. Label the event Alfonso got ticket 14 by $A_{14}$ and Colin got ticket 23 by $A_{23}$. Then by the addition rule
$$\begin{gathered} Pr(\text{ (}{A}_{14}\text{ first and }{C}_{23}\text{ second) or (}{C}_{23}\text{ first and }{A}_{14}\text{ second})) \\ =Pr(A_{14})\times <!--\; \times \; -->Pr(C_{23}\mid <!--\; \mid \; -->A_{14})+Pr(C_{23})\times <!--\; \times \; -->Pr(A_{14}\mid <!--\; \mid \; -->C_{23})=\frac{1}{50}\times <!--\; \times \; -->\frac{1}{49}\times <!--\; \times \; -->2. \end{gathered}$$
I realize that once the tickets are sold, then only one of $\{A_{14}{C}_{23},C_{23}{A}_{14}\}$ must occur, but before the tickets are sold both possibilities are plausible. Why would the probability change before and after the tickets are sold.