I'm conceptually/intuitively confused about when we would use the joint probability (multiplication rule) versus a conditional probability, specifically when the dependent event has no chance of occurring without the other event. Here's an example that illustrates my confusion: Suppose we have a weird city, where there's a 20 percent chance of rain (so a 80% chance of any other weather). Now suppose that it can only rain frogs when it rains, and there's a 10% chance of frogs. Why can we not use bayes rule to model: $$Pr(frogs | rain) = \frac{Pr(rain | frogs)\cdot Pr(frogs)}{Pr(rain)} = \frac{1 \cdot 0.1}{0.2}$$ I know the correct way of doing it would be the multiplication rule, but why can we not use a conditional probability? We know that the probability of raining frogs is 0.1 because it cannot occur in any other scenario other than when it rains, and if there are raining frogs, we must have rain so $Pr(rain | frogs)=1$. And we also have $Pr(rain)=0.2$, so it seems like we have everything we need for conditional probability. The correct answer would of course be $0.2\cdot0.1=0.02$. What am I missing aside from the nonsensical results? Edited for sensical numbers, but assume that the probability of raining frogs is just the proportion of days on which it rains.
123256
When talking about events, there is the following formula called Bayes' rule, where $A$ and $B$ are random events: $$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$$  Now let's say that for now only $A$ happened. I suppose that $P(B|A)$ is called posterior probability (beside conditional probability) since there is an evidence, $A$.  But is $P(A|B)$ also called posterior probability?   Edit: An (artificial) example:  $A$ = "Last year I had no accidents."  $B$ = "This year I will have no accidents."  Edit 2:  Simply put: $P(\text{something}|\text{evidence})$ = posterior probability  But $P(\text{evidence}|\text{something})$ = posterior?  Doesn't "posterior" have something to do with the chronology of events (their order of happening in time)?
93796
Let $A$ and $B$ be two events of a probability space. Prove that  $\displaystyle|P(A\cap B)-P(A)P(B)|\leq \frac{1}4$   I think it's a very challenging problem, and I've made no progress so far ...  Can someone give me a hint ?
40871
Probability that $X \ge Y$ given N rolls on a $d20$  Example:  Originally, we wanted to roll a $13$ or higher $7$ times in a row. In and attempt to improve the odds and make it easier to understand we changed it to be $7(N)$ $d20s$ where one must be greater than or equal to $91(Y)$. So, what is the probability that $X$(sum of rolls) is greater than or equal to $Y(91)$ given $N(7 rolls)?$   We think the answer is $0.16%$ or $1/625$ because in order to achieve the minimum of $91$, one must roll at least a $13$ on every roll. What we're wondering is does the fact that it's now cumulative change the probability?  Thanks for your time.
84249
A coin is tossed three times. The probability of zero heads is 1/8 and the probability of zero tails is 1/8. What is the probability that there is at least one head and at least one tail?  So, if P(zero heads)= 1/8 , then that should be the same of p(all tails)?  We would use the complement rule and Multiplication Rule?   P(at least one head) = 1 - P(no heads) = 1 - 1/8= 7/8 P(at least one tail) = 1 - P(no tails) = 1 - 1/8= 7/8  would I multiply the two values?
30972
Note: I am aware of  differences between Bayesian and frequentist statistics. For example &quot;what is the probability that the coin on the table is heads&quot; doesn't make sense in frequentist statistics, since it has either already landed heads or tails -- there is nothing probabilistic about it. So the question has no answer in frequentist terms. But such a difference is specifically not the kind of difference I'm asking about. Rather, I would like to know how their predictions for well-formed questions actually differ in the real world, excluding any theoretical/philosophical differences such as the example I mentioned above. So in other words: What's an example of a question, answerable in both frequentist and Bayesian statistics, whose answer is different between the two? (e.g. Perhaps one of them answers &quot;1/2&quot; to a particular question, and the other answers &quot;2/3&quot;.) Are there any such differences?  If so, what are some examples?  If not, then when does it actually ever make a difference whether I use Bayesian or frequentist statistics when solving a particular problem? Why would I avoid one in favor of the other?
64910
I'm trying to wrap my head around the result of Bayes Theorem applied to the classic mammogram example, with the twist of the mammogram being perfect.  That is,  Incidence of cancer: $.01$  Probability of a positive mammogram, given the patient has cancer: $1$  Probability of a positive mammogram, given the patient does not have cancer: $.01$  By Bayes:  P(cancer | mammogram+) = $\dfrac {1 \cdot .01}{(1 \cdot .01) + (.091 \cdot .99)}$  $ = .5025$  So, if a random person from the population takes the mammogram, and obtains a positive result, there is a 50% chance they have cancer? I'm failing to intuitively understand how the tiny 1% chance of a false positive in 1% of the population can trigger a 50% result. Logically, I would think a perfectly true positive mammogram with a tiny false positive rate would be much more accurate.
40005
I'm self-studying a textbook () (not for course credit) and wanted to post my solution to a problem related to conditional probability.   Given:  $$ \begin{align} &amp;P((A \cup B)^C) = 0.6 \\ &amp;P(A \cap B) = 0.1 \\ &amp;E = A\ xor\ B\ \text{(A or B but not both)} \end{align} $$  Find $P(E\ |\ A \cup B)$  Below is my solution. Is this correct or else where did I go wrong?   $$ \begin{align} P(E\ |\ A \cup B) &amp;= \frac{P(E\ \cap\ (A \cup B))}{P(A \cup B)} \tag 1 \\  \\ P(E\ \cap (A \cup B)) &amp;= P((E \cap A)\ \cup\ (E \cap B)) \tag 2 \\ \\ &amp;= P((A \cup B) \cup (A \cup B)) \\ \\ &amp;= P(A \cup B) \\ \\ \therefore \\ \\ P(E\ |\ A \cup B) &amp;= \frac{P(A \cup B)}{P(A \cup B)} = 1 \tag 3 \\ \\ \end{align} $$  Noting that $E \cap B$ = [(A xor B) or B] = $A \cup B$ and similarly $E \cap A$ = [(A xor B) or A] = $A \cup B$.
66158
First, I apologise for the vague title. I couldn't think of a short way to represent the problem. Also, I am aware that a similar question exists, but I have a little bit more insight.  The problem is:      A man is known to speak the truth three out of four times. He throws a   die and reports that it is a six. Find the probability that it   actually is a six.   The answer in the book is $ \frac{3}{8} $, which sounds completely off to me, given the numbers, on first glance. I decided to try it on my own. After some contemplation, it indeed is true that $ \frac{3}{8} $ is correct. But, the question's vagueness makes it interesting. It's sort of a naive answer!  First, here's the book's solution:  Let $ E $ be the event that the man reports that a six occurs, $ S_1 $ be the event that six occurs and $ S_2 $ be the event that six does not occur.  $ P(S_1) $ = Probability that six occurs = $ \frac{1}{6} $  $ P(S_2) $ = Probability that six does not occur = $ \frac{5}{6} $  $ P(E~|S_1) $ = Probability that the man reports that a six occurred when it has actually occurred (truth) = $ \frac{3}{4} $  $ P(E~|S_2) $ = Probability that the man reports that a six occurred when it hasn't actually occurred = $ \frac{1}{4} $  $ P(S_1|E) $ = Probability that a six has actually occurred when the man claims so =  $$ \frac{P(S_1)~P(E~|S1)}{P(S_1)~P(E~|S_1)+P(S_2)~P(E~|S_2)} = \frac{3}{8} $$  Link to original: Page 25 of  (link updated 22/11/2015; tends to break often).  Here's the first of my two solutions:    The question is a little vague and it doesn't tell us what the man says when he doesn't get a six. Assuming he lies every time he doesn't get a six (which is not quite what the question says), would bring up $ P(E~|S_2) $ to $ 1 $, which skyrockets $ P(lie) $ to $ \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6}.\frac{1}{4} = \frac{7}{8} $.  Also, $ P(E~|S_1) = \frac{3}{4} $ and $ P(E) = \frac{5}{6} + \frac{1}{6}.\frac{3}{4} = \frac{23}{24} $.  This brings us to, $$ P(S_1|E) = \frac{P(S_1)P(E~|S_1)}{P(E)} = \frac{3}{23} = 13.04\% $$  Please note that I don't care about the sixes rolled when he lies about it being a six, because that condition is not applicable to the question.  I'm very skeptical about my calculations! I decided to whip up a simple program and see what the probability distribution is for $$ \frac{Number~of~sixes~rolled~while ~saying~it's~a~six}{Number~of~dice~rolls} $$  Here's the result:    Considering the book says $ P(E~|S_2) = \frac{1}{4} $, which is in essence is contradictory to the question, let's also consider the other possibility:    It's even more vague in this scenario! If the man lies after any given roll, there is a $ \frac{1}{5} $ probability that he will lie about it being any other number. This creates a new branch at each lie, which splits into five. In this case,  $$ P(E) = \frac{1}{6}.\frac{1}{4}.\frac{1}{5}.\frac{5}{1}+\frac{1}{6}.\frac{3}{4} =  \frac{1}{6} $$  $ P(E~|S_1) $ still remains $ \frac{3}{4} $, but $ P(E~|S_2) = \frac{1}{4}.\frac{1}{5} = \frac{1}{20} $.  $$ P(S_1|E) = \frac{P(S_1)P(E~|S_1)}{P(E)} = \frac{3}{4} = 75\% $$  This seems like the more intuitive answer, and it is why I doubted the original answer in the first place. Given a truth probability of $ \frac{3}{4} $, there was no way that the probability of a six was $ 37.5\% $!    The ambiguity of the question is what makes $ \frac{3}{8} $ wrong to me. Entirely out of curiosity, which answer fits best with the question and are these calculations correct or baloney?
82231
I've just read in a stochastics textbook:     Let $(\Omega, P)$ be a discrete probability space.      (a) The events $A_i \subseteq \Omega, i=1,2, \dots$ are called independent, if   $$P(A_{i_1} \cap A_{i_2} \cap \dots \cap A_{i_k}) = P(A_{i_1}) \cdot \dots \cdot A_{i_k}$$   for all finite sets $\{i_1, \dots, i_k\} \subsetneq \{1,2, \dots\}$ and all $k \geq 2$.      (b) The events $A_i \subseteq \Omega, i=1,2,\dots$ are called pairwise independent, if   $$P(A_{i_1} \cap A_{i_2}) = P(A_{i_1}) \cdot P(A_{i_2})$$   for all pairs $\{i_1, i_2\} \subsetneq \{1,2,\dots\}$.   After these two definitions it states that pairwise independent events are not always independent.  Do you have an example?
101574
I'm stuck on the following probability problem: parents keep having children until they have one girl, at which point they stop; and babies are girls with probability 0.49. If we select a child uniformly at random (from the entire population of children) what's the probability he or she has exactly one sibling?  If a couple has only $2$ children it means that they had first a boy and then a girl. So why the probability asked is not $0.51 \cdot 0.49$? What I am missing?   It seems that the correct answer is $2 \cdot 0.51 \cdot (0.49)^2$, but I don't know how to get this result. Thanks for your help.
57418
I recently read a math problem and, having not yet taken anything beyond calculus 1, was curious about how to solve it correctly.   Problem:  Calculate the probability of two people meeting at the same location between 1 and 2 p.m. Assume both people show and person 1 will wait 15 minutes for person 2.     Doesn't this probability increase as the time frame shrinks?  For example, Isn't there a much smaller chance the pair will meet if person 1 arrives at 1:00 vs arrives at 1:45 (the probability of meeting becomes 1).  Is this somewhat easily answered using basic statistics?  Just curious if my thought process is accurate or way wrong.
35945
We had this question arise in class today and I still don't understand the answer given. We were to assume that drawing cards are independent events. We were asked what the probability that the second card drawn is a queen if we take two from the deck. The answer given was 4/52, which seems counter-intuitive to me. How is the probability still 4/52 if there was a card drawn before it? What if the first card drawn was a queen?
82881
Before reading the Wikipedia article, my idea to calculate the probability was as follows:  $1-\left(\frac{364}{365}\right)^{_nC_{2}}$  Basically, I thought to compare all combinations of pairs ($_nC_{2}$) and require them to have different birthday ($\frac{364}{365}$). It gave a very similar result to the wiki method, but is slightly different and I can't tell how  my way of thinking is wrong.  Can anyone tell me what I'm missing?
52202
According to Bayes' theorem, $P(y|\theta)P(\theta) = P(\theta|y)P(y)$. But according to my econometric text, it says that $P(\theta|y) \propto P(y|\theta)P(\theta)$. Why is it like this? I don't get why $P(y)$ is ignored.
72516
Having read a little about exchangeability, I went back to thinking about the iid condition required for the central limit theorem. It struck me that if two random variables are drawn from an identical distribution, the occurrence of one event does not make it more or less probable for the occurrence of another event. I have no doubts that I am wrong and there is a reason why both independence and identical distribution are required. I just don't know why!
91646
I am confused about how to proceed with solving 3rd and 4th question.  The question is :  Alice has two coins. The probability of Heads for the first coin is  ${1\over4}$, and the probability of Heads for the second is ${3\over4}$. Other than this difference, the coins are indistinguishable. Alice chooses one of the coins at random and sends it to Bob. The random selection used by Alice to pick the coin to send to Bob is such that the first coin has a probability  $p$  of being selected. Assume that  $0&lt;p&lt;1$. Bob tries to guess which of the two coins he received by tossing it  $3$  times in a row and observing the outcome. Assume that for any particular coin, all tosses of that coin are independent.  1. Given that Bob observed  $k$  Heads out of the  $3$  tosses $(where                     k=0,1,2,3 )$, what is the conditional probability that he received the first coin?  My answer = ${3^{3-k}⋅p \over3^{3-k}⋅p+3^k⋅(1−p)}$  2. We define an error to have occurred if Bob decides that he received one coin from Alice, but he actually received the other coin. He decides that he received the first coin when the number of Heads,  $k$ , that he observes on the  $3$  tosses satisfies a certain condition. When one of the following conditions is used, Bob will minimize the probability of error. Choose the correct threshold condition.  My answer = $k ≤ {3\over2}+{1\over2}log_3 {p\over1−p}$  3. For this part, assume that  $p={3\over4}$  3 a. What is the probability that Bob will guess the coin correctly using the decision rule from part  2?  3 b. Suppose instead that Bob tries to guess which coin he received without tossing it. He still guesses the coin in order to minimize the probability of error. What is the probability that Bob will guess the coin correctly under this scenario?  4. Bob uses the decision rule of Part 2. If  $p$  is small, then Bob will always decide in favor of the second coin, ignoring the results of the three tosses. The range of such  $p's$ is  $[0,t)$. Find $t$.
96887
I'm trying to figure out what is the probability with 3 D10's if I were to roll them one after another."001" is one and "000" is a thousand. Would every number have the same probability or doubles and triples be less common?  And please explain.