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3

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}\\ ...


1

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:\space$sample space:$\space$all 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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