So I'm having this assignment in probability regarding conditional probability that states the following:  Each of 25 exams papers has 2 questions written on it. Neither of the 50 questions repeats itself. The student knows the answer for 44 questions. In order for the student to pass the exam he must answer correctly either on the two questions for the paper he has chosen or on one question on the first paper he has chosen and on the first question on the second paper he has chosen.  What is the probability that the student will pass the exam?  Any ideas?  Any sort of help will be appreciated.  Thank you in advance!
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There're $25$ tests, each with $2$ questions. There're total of $50$ different questions (the questions do not repeat). The student knows the answer of $44$ questions. To pass the exam, the student needs to correctly answer both questions on the first test he drew, or answer $1$ question from the test he drew first, and $1$ question from the test he drew second. (meaning he draws twice if he answers a question incorrectly on his first try)  What is the probability that the student will pass the exam?  I have never faced a problem like this one so any help is appreciated. I was thinking about approaching it "manually" with some applied combinatorics but I feel like there's a much easier way that I am not familiar with.
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A student is applying to a PhD program in Computer Sciences at Harvard and MIT. He estimates that he has two out of three chances of being accepted at Harvard and two out of five chances at MIT. He further estimates the probability that he will be accepted by both has probability of one to five. What is the probability that he is accepted by MIT if he is accepted by Harvard? Is the event "accepted by MIT" independent of the event "accepted by Harvard".  I don't understand if these are odds or probabilities
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Just moments ago I stumbled upon the definition of conditional probability and I'm experiencing some difficulties in acquiring an intuitive understanding of why it's defined the way it is.   $$P(E|F)= \dfrac{P(E \cap F)}{P(F)}$$  I'm more familiar with taking into consideration the amount of outcomes of interest and dividing that amount by the amount of outcomes in the sample space. I don't know how to interpret the logic behind conditional probability. I understand that $P(F)$ is the probability that the event which is known to have occured occurs. So I suppose that it can be calulated the way I am familiar with. Presumably the same goes for $P(E \cap F)$ by using the Inclusion–exclusion principle. But why one is divided by the other I cannot fathom. Why is it defined this way? How does one make sense of it.
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Ok, I know this is wrong, but I want someone to tell me why.   Let's take a normal heads tails example of a fair coin. The probability of getting head = 1/2. And I write this is because, either it will be heads, or not. Hence two cases that makes it 1/2.  Now, I know I can't do this to all the cases. For example, the probability of getting a 2 when I roll a dice. I know the answer is 1/6 but why can't I do, either the outcome will be 2 or not. And in that case, my probability is 1/2.
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I'm in a computer algorithms course and have a question about basic probability. My math background includes no more than discrete math and a little calculus, so this probability question left me completely stumped. I checked for similar questions but didn't see any. If this question has been answered already I apologize in advance but I don't even know the correct terms to search to find an answer.  The question goes like this:   Suppose you're playing a set of games and the first to win 10 games  wins the set. Each player has an even chance (1/2) of winning a  game. What is the probability that the first player will win if she  has already won i games and the second player has already won j  games? How do you calculate this for different values of i and j?    Please word the explanation for a non-maths student. I need to provide a recursive formulation for a function that computes this probability then solve it using dynamic programming. No worries on recursion or dynamic programming. I can solve that once I understand how to solve for probability.
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I've been grading exams for most of the day. Once I finished grading, I started entering the grades into my gradebook -- one by one, from top to bottom on the stack. About halfway through, I entered one students grade and the next student on the stack was also the next person alphabetically in the gradebook. What is the probability of this happening with $n$ students, all of whom have unique names?  Equivalent question:  For a random permutation $\left(a_1,a_2,\ldots,a_n\right)$ of the list $\left(1,2,\ldots,n\right)$, what is the probability that there exists at least one entry $k$ of the permutation that is followed immediately by $k+1$ (that is, $k = a_i$ and $k+1 = a_{i+1}$ for some $i \in \left\{1,2,\ldots,n\right\}$) ?  For small $n$, it's not hard to exhaustively calculate the probability. But my combinatorics skills are rusty, and I don't think I can easily calculate this for my 30 students.
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Assume that there are an arbitrarily large number of copies of ten diﬀerent discrete math books (which we will call titles) for cmsc250. Each student must choose one title as the course text book. Assume that the ﬁve hundred students taking the course are indistiguishable.  How many diﬀerent ways can the students choose titles, if Epp must be chosen by at most twenty students?   I understand that this is a stars and bars problem where the formula would be ${n+r-1}\choose{r-1}$, but whats throwing me off is the "at most" parameter of the problem. My answer is ${480+9-1}\choose{9-1}$ My reasoning was that we now have one less "bar", and that should be shown by lowering $r$ by 1  Am I approaching this the right way?
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I have the following problem: A class has 20 students, 16 females, 4 males. Find all possibilities of choosing 6 students so that at least one is male.  I did the following, pick one male from the 4, you have 4 possibilities, then you are left with 19 students, pick 5, so I get 4*${19}\choose{5}$ possibilities.  However, one can notice that if you want no one to be male, then the possibilities are ${16}\choose{6}$, so, one could argue that the solution is instead ${20}\choose{6}$ - ${16}\choose{6}$.  So, which is solution is correct, and why?  $$4* {{19}\choose{5}}$$  or $${{20}\choose{6}} - {{16}\choose{6}}$$
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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 $t$.
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A pair of dice is rolled repeatedly until each outcome (2 through 12) has occurred at least once. What is the expected number of rolls necessary for this to occur?  Notes: This is not very deep conceptually, but because of the unequal probabilities for the outcomes, it seems that the calculations involved are terribly messy.  It must have been done already (dice have been studied for centuries!) but I can't find a discussion in any book, or on line. Can anybody give a reference?
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I was watching the movie 21 yesterday, and in the first 15 minutes or so the main character is in a classroom, being asked a "trick" question (in the sense that the teacher believes that he'll get the wrong answer) which revolves around theoretical probability.  The question goes a little something like this (I'm paraphrasing, but the numbers are all exact):  You're on a game show, and you're given three doors. Behind one of the doors is a brand new car, behind the other two are donkeys. With each door you have a $1/3$ chance of winning. Which door would you pick?  The character picks A, as the odds are all equally in his favor.  The teacher then opens door C, revealing a donkey to be behind there, and asks him if he would like to change his choice. At this point he also explains that most people change their choices out of fear; paranoia; emotion and such.  The character does change his answer to B, but because (according to the movie), the odds are now in favor of door B with a $1/3$ chance of winning if door A is picked and $2/3$ if door B is picked.  What I don't understand is how removing the final door increases the odds of winning if door B is picked only. Surely the split should be 50/50 now, as removal of the final door tells you nothing about the first two?  I assume that I'm wrong; as I'd really like to think that they wouldn't make a movie that's so mathematically incorrect, but I just can't seem to understand why this is the case.  So, if anyone could tell me whether I'm right; or if not explain why, I would be extremely grateful.
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I have a pretty basic probability question, but I'm just having difficulties remembering what distribution this is.  A coin is tossed until a head appears two times in a row. Given that we are using a fair coin, what is the probability that we toss the coin exactly 4 times such that the two consecutive heads are the 3rd and 4th trials?  I know that this would be fairly easy to solve by brute force and listing out the different probabilities, but I'm a little confused as to how to use a "shortcut" such as identifying this as a binomial distribution. I also don't know how to account for the probability that we are using 4 tosses.  Any help is appreciated. Thanks!
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A teacher wants to randomly form two teams of 5 students from a group   of 5 girls and 5 boys for a sports activity. Two of the girls, Ann and   Alice, are selected as team leaders. Find the probability that one   team has exactly 3 girls.   If i do this way $$\dfrac{{3 \choose 2}{5\choose 2}{1\choose 1}{3\choose 3} \times 2!}{{8 \choose 4}{4 \choose 4}} = \dfrac{6}{7}$$  Which turns out to tally with the answer. But i got one big question, is my way correct. Do i really need to times $2$ here? Please see my link below to my previous question cause the previous question make me doubt if i should times $2$.
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I have the following question:  "We draw one card at a time without replacement from the top of a shuffled standard poker deck and stop when we draw an ace. Let $X$ be the number of cards we have drawn, then calculate the probabilities for $X=10$, $X=50$, and $X &lt; 10$."  I have a theory as to how to solve the first one: calculate the probability of the drawing 10 cards WITHOUT getting an ace ($\binom{48}{10} / \binom{52}{10}$) and later subtracting this from 1. For the other two, I am a little bit lost.  Regarding the second one, I think that if I draw 50 cards from the deck, the probability of drawing an ace should be 1, shouldnt it?  And unfortunately I'm lost on the last part. I would love a clue on this matter.  Thanks
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I have a simple question regarding "conditional probability" and "Likelihood". (I have already surveyed this question  but to no avail.)  It starts from the Wikipedia . They say this:     The likelihood of a set of parameter values, $\theta$, given   outcomes $x$, is equal to the probability of those observed outcomes   given those parameter values, that is      $$\mathcal{L}(\theta \mid x) = P(x \mid \theta)$$   Great! So in English, I read this as: "The likelihood of parameters equaling theta, given data X = x, (the left-hand-side), is equal to the probability of the data X being equal to x, given that the parameters are equal to theta". (Bold is mine for emphasis).  However, no less than 3 lines later on the same page, the Wikipedia entry then goes on to say:     Let $X$ be a random variable with a discrete probability distribution   $p$ depending on a parameter $\theta$. Then the function      $$\mathcal{L}(\theta \mid x) = p_\theta (x) = P_\theta (X=x), \, $$      considered as a function of $\theta$, is called the likelihood   function (of $\theta$, given the outcome $x$ of the random variable   $X$). Sometimes the probability of the value $x$ of $X$ for the   parameter value $\theta$ is written as $P(X=x\mid\theta)$; often   written as $P(X=x;\theta)$ to emphasize that this differs from   $\mathcal{L}(\theta \mid x) $ which is not a conditional probability,   because $\theta$ is a parameter and not a random variable.   (Bold is mine for emphasis). So, in the first quote, we are literally told about a conditional probability of $P(x\mid\theta)$, but immediately afterwards, we are told that this is actually NOT a conditional probability, and should be in fact written as $P(X = x; \theta)$?   So, which one is is? Does the likelihood actually connote a conditional probability ala the first quote? Or does it connote a simple probability ala the second quote?  EDIT:  Based on all the helpful and insightful answers I have received thus far, I have summarized my question - and my understanding thus far as so:    In English, we say that: "The likelihood is a function of parameters, GIVEN the observed data." In math, we write it as: $L(\mathbf{\Theta}= \theta \mid \mathbf{X}=x)$. The likelihood is not a probability.  The likelihood is not a probability distribution.  The likelihood is not a probability mass. The likelihood is however, in English: "A product of probability distributions, (continuous case), or a product of probability masses, (discrete case), at where $\mathbf{X} = x$, and parameterized by $\mathbf{\Theta}= \theta$." In math, we then write it as such: $L(\mathbf{\Theta}= \theta \mid \mathbf{X}=x) = f(\mathbf{X}=x ; \mathbf{\Theta}= \theta) $ (continuous case, where $f$ is a PDF), and as $L(\mathbf{\Theta}= \theta \mid \mathbf{X}=x) = P(\mathbf{X}=x ; \mathbf{\Theta}= \theta) $ (discrete case, where $P$ is a probability mass). The takeaway here is that at no point here whatsoever is a conditional probability coming into play at all.  In Bayes theorem, we have: $P(\mathbf{\Theta}= \theta \mid \mathbf{X}=x) = \frac{P(\mathbf{X}=x \mid \mathbf{\Theta}= \theta) \ P(\mathbf{\Theta}= \theta)}{P(\mathbf{X}=x)}$. Colloquially, we are told that "$P(\mathbf{X}=x \mid  \mathbf{\Theta}= \theta)$ is a likelihood", however, this is not true, since $\mathbf{\Theta}$ might be an actual random variable. Therefore, what we can correctly say however, is that this term $P(\mathbf{X}=x \mid \mathbf{\Theta}= \theta)$ is simply "similar" to a likelihood. (?) [On this I am not sure.]   EDIT II:  Based on @amoebas answer, I have drawn his last comment. I think it's quite elucidating, and I think it clears up the main contention I was having. (Comments on the image).    EDIT III:   I extended @amoebas comments to the Bayesian case just now as well:
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Hey first I want to start of by saying that this is homework. I'm just looking for guidance. These problems involve using a deck of cards.  I want to find the conditional property that the second card will be an ace given that the first is a heart. I did this by by trying to find P(B/A)   A: heart was the first card B: drawing the ace as the second card P(B/A) = P(A n B) / P(A)          I know that the probability of getting a heart as the first          card and the ace on the second is 1/52 from a previous problem        = (1/52)/(13/52)        = (1/13) P(B/A) = (1/13)   Now the second portion of the question asks to find the probability that the second card draws an ace using the law of total probability. I think this is how you implement it  A: Second card is an ace B: First card is not an ace C: First card is an ace P(A) = P(A|C)P(C) + P(A|B)P(B) P(A|C)P(C) = (((4 choose 1)(3 choose 1))/(52 choose 2)) * (4/52) P(A|B)P(B) = (((48 choose 1)(4 choose 1))/(52 choose 2)) * (48/52) P(A) = 386/2783   However the last part of the question asks if they are independent or not, and I know that they are only independent if   A: First card is a heart B: Second card is an ace P(B|A) = P(B)   And after the above calculations they are not independent. However after some further thought I feel like this might be wrong. I think that whether the first card is a heart should be independent of whether you choose an ace as your second draw. If someone could deny or confirm this, it would be great.
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Mr. Jones has three shirts: Red, Green and White. Each day he picks randomly a red shirt with probability of $\frac{1}{2}$, green with probability of $\frac{1}{3}$ and white with probability of $\frac{1}{6}$.     What is the probability that he wears all three shirts after 6 days?   My try:   $$\binom{6}{1}\binom{5}{1}\binom{4}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}\cdot[\binom{3}{2}(\frac{1}{2})^2\cdot[\frac{1}{3}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{3})^2\cdot[\frac{1}{2}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{6})^2\cdot[\frac{1}{2}+\frac{1}{3}]+\binom{3}{1}\binom{2}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}]$$  First, we pick possessions for red, green and white. Then, we multiply it by the other combinations possible of the other three possessions left, and summing it up (I didn't count all possibilities). The problem is the answer is bigger than 1, so it must be wrong. Also, I guess there is more elegant way.  Thanks in advance.