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Suppose $i$ is an integer between $1$ and $24$, and let $A_i$ be the event that you roll an $i$ at least $4$ times in ten rolls. Notice that two of the $A_i$ could occur, but not three, since that would be twelve different rolls. By the inclusion/exclusion principle, you get that $$\mathbb{P}(\cup A_i) = \sum \mathbb{P}(A_i) - \sum_{i\ \neq j}\mathbb{P}(A_i ... 2 HINT: Since the company opens at most one store in a block, if a town has n blocks, we can represent any possible way of opening stores by a string of n zeroes and ones: a 0 represents a block in which the company does not open a store, and a 1 represents a block in which the company does open a store. The other restriction means that we want only ... 2 The problem is equivalent to Coupon collector's problem. The expected value is$$\mathbb{E}(X) = N H_N \approx N \, ln \, N$$where H_N is N-th harmonic number. Here \mathbb{E}(X) \approx 2364.64 The idea in solving this problem is calculating the expected number of people such that the number of different birthdays we have written increases ... 2 P(X=5)=\dfrac{\binom{4}{4}}{\binom{100}{5}} P(X=6)=\dfrac{\binom{5}{4}}{\binom{100}{5}} P(X=7)=\dfrac{\binom{6}{4}}{\binom{100}{5}} \dots And in general:$$\forall{n\in[5,100]}:P(X=n)=\dfrac{\binom{n-1}{4}}{\binom{100}{5}}$$In words: Take ball #n, and choose another 4 balls out of balls #1,\dots,n-1. 2 Count how many ways n can be the largest number. If you replace the balls, there are n^5 ways they can be \leq n, minus (n-1)^5 ways they are all less than n. If you don't replace the balls, there are n-1\choose4 ways that the largest is n. 2 Using the formula for conditional probability:$$p(\text{fair}\mid\text{head})=\frac{p(\text{fair and head})}{p(\text{head})}=\frac{1/4}{3/4}=1/3$$2 Let A be the event : "The coin is fair" Let B be the event : "Heads appears." We have P(B)=\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times 1=\frac{3}{4} and P(A\cap B)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4} So, we have P_B(A)=\frac{1}{3} 2 Regardless of whether or not Plura is at the gym, Carla has a 20% chance of showing up at the gym. Therefore, the probability that Plura will meet Carla at the gym is 20%. Similarly, the probability that Plura shows up at the gym is 15%, so the probability that Carla meets Plura at the gym is 15%. The probability that Plura AND Carla show up is ... 2 Let X be the event of a faulty element (and X_n the n^{th} element being faulty). And let A and B be the event of picking both elements from the manufacture A and B, respectively. We look for P(X_2\mid X_1). Since nothing's said, let's assume that all events are independent, since otherwise we miss information. Note that since A and B are ... 1 Define events:$$F_1 = \text{"first product is faulty"} \\ F_2 = \text{"second product is faulty"} \\ A = \text{"products chosen from $A$"} \\ B = \text{"products chosen from $B$"}.$$Then, we require \begin{eqnarray*} P(F_2\mid F_1) &=& \dfrac{P(F_2\cap F_1)}{P(F_1)} \\ && \\ &=& \dfrac{P(F_2\cap F_1 \mid A)P(A) + P(F_2\cap F_1 ... 1 Hint: Satisfaction of condition X_2=m means that a number in \{1,2,3,6\} was thrown exactly n-m times. What is the probability that k of these times it was a number in \{1,2,3\}? 1 All of these statements can be false. What follows is more or less the standard counterexample of a local martingale that is not a martingale. I've stolen the details from an MO post of mine which constructs something slightly different. (To avoid search-and-replace I'm keeping my process called Y instead of \beta.) Set T=1. Let r(t) be any ... 1 \newcommand{\E}{\operatorname{E}}If U, V are undependent and \E U and \E V both exist then \E(UV)=(\E U)(\E V). Therefore$$ \E\Big((X_i-\E X_i)(X_j-\E X_j\Big) = \E(X_i-\E X_i) \E(X_j-\E X_j) $$and$$ \E(X_i-\E X_i) = \E X_i - \E(\E X_i)) = \E X_i - \E X_i=0 and similarly for j. \begin{align} & \sum_{k=2}^\infty k(k-1)q^{k-1}p = qp ... 1 One way to approach it is as follows. Firstly, Y is either X_2 or X_3 and with equal probability because f_{X_1,X_2,X_3} is symmetric wrt those two variables. So we begin by conditioning on the event that X_2=Y: \begin{eqnarray*} E(X_2\mid X_1=x, Y=y) &=& E(X_2\mid X_1=x, Y=y, X_2=y)P(X_2=y\mid X_1=x, Y=y) + \\ && E(X_2\mid X_1=x, ... 1 I assume that (abc)(def)(ghi)(jkl) is considered the same result as (def)(ghi)(jkl)(abc) (i.e. rotations of tables is irrelevant) and (bca)(def)(igh)(klj) (i.e. rotations within tables is irrelevant), but is considered different from (bac)(def)(ghi)(jkl) (mirrors of tables considered different) and is considered different from (def)(abc)(ghi)(jkl) ... 1 For i=1,\dots,N let X_{i} take value 1 if number i is not drawn and value 0 otherwise. Then X:=X_{1}+\cdots+X_{N} equals the number of numbers that are not drawn. Then \mathbb{E}X=\mathbb{E}\left(X_{1}+\cdots+X_{N}\right)=\mathbb{E}X_{1}+\cdots+\mathbb{E}X_{N}=\mathbb{P}\left(X_{1}=1\right)+\cdots+\mathbb{P}\left(X_{N}=1\right) An expression ... 1 The sample space with the (product) probabilities assigned to the elementary events: \begin {matrix} \text{#}&\text{bus}&\text{train1}&\text{train2}&\text{prod. of probs.}&\text{resulting prob}\\ 1&\text{O}&\text{O}&\text{O}&\frac{1}{3}\frac{3}{4}\frac{3}{4}&\frac{9}{48}\\ ...

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I think it is clearer to break up 1st and 2nd rolls into separate entities: $i)1st:$ sample space:${6\choose 1}=6$. Probability that is $4$:$\space {|(4)|\over 6}={1\over 6}.\space$$2nd:\spacesample space:\spaceall possible 1st rolls combined with all possible 2nd rolls:\space {6\choose 1} {6\choose 1}=6*6=36. Probability that adds up to ... 1 Let W be the number of tagged fish caught. Then the number of untagged fish caught is 5-W. Net earnings Y are given by$$Y=10W+2(5-W)-25=8W-15.$$It follows that \text{Var}(Y)=8^2\text{Var}(W). I have not checked the correctness of your calculation of \text{Var}(W). Note that the unit of variance is fishes^2. 1 The number of permutations without fixed points in S_5 is given by:$$ 5!\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}\right)=\color{red}{44}\tag{1} $$by the inclusion-exclusion principle, hence the wanted probability is:$$ \frac{44}{4^5} = \color{red}{\frac{11}{256}}\approx 4,3\%.\tag{2}$\$

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