# Tag Info

0

The probabilities should not sum up to $\frac1{\sqrt N}$. Contemplate the problem with "square" replaced with "even", where you should have $p=1$ or $p=0$, depending on the unit digit being even/odd, whereas the overall probaility is $\frac12$. Instead, your $K$ should be just $\frac1{\sqrt N}$, do you see why?

0

@Stones Yes, but the table is not quite right. If there are three or more 1s or 2s, then it does not matter how many 6s there are. So N5 the number of ways of getting five 1s or 2s is 32. Similarly, N4 the number of ways of getting four is $5\times 2^4\times 4=320$ and N4 the number of ways of getting three is $10\times 2^3\times 4^2=1280$. If we get just ...

2

$${n\choose n-j-1}F(z_{k-1})^{n-j-1}{j+1\choose j}(F(z_k)-F(z_{k-1}))^j$$ Deciphering the formula: The first binomial factor is the number of ways of choosing $n-j-1$ elements from the whole sample (the ones that will be below $z_{k-1}$). The first power of $F$ is the probability that this given subsample is indeed below $z_{k-1}$. The second binomial ...

0

First, you wrote that if $X_n$ is not constant a.s., then $\forall c\in\mathbb{R},\epsilon>0; P(|X_n-c| > \epsilon) > 0$. This is not true. Change the $\forall$ to an $\exists$ and it will be true. Here's one approach: $\implies)$ So $X_n$ is not constant a.s. That means there exist $c\in\mathbb{R}$ and $\epsilon>0$ such that ...

2

With a nod to @TonyK that this real world problem is impossible, I will treat it as a pure mathematical question. Before the game was played: $$P(A)=P(B)=P(C)=0.2$$ and $$P(D)=0.4$$ It was a given that in order to win, Team D must win its semi and would play either A or B in the final. Assuming that its chances against either are the same, what has changed ...

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Thanks for a detailed explanation @vadim123. for the said strategy where the multiplier is 2, if we start from $2 instead of 1 so that we would increase the winning by 2 and decrease the required winning times from 50 to 25 - more risky; we still have 0.938 (15/16) chance of winning PER SEQUENCE. From your example above, we have 0.984 (63/64) chance of ... 0 This is essentially a counting problem: in the$6^5=7776$possible outcomes, how many satisfy the condition "the number of rolls resulting in 1 or 2 is greater than the number of rolls resulting in 6"? The probability is this number divided by$6^5$. We can find this number by filling five initially empty cells with the possible numbers. Hint: We can find ... 0 We want to know the probabilities that A, C, or D will win, given that we now know that B will not. We shall use Bayes' Theorem to calculate the conditional probabilities.$\begin{align} P(D \mid \neg B) & = \dfrac{P(D\cap \neg B)}{P(\neg B)} \\ & = \dfrac{P(D)}{P(A)+P(C)+P(D)} & \because \text{mutually exclusive events} \\ & = ...

5

A simple way is to suppose that person A has been assigned to Group 1. Note that $3$ of the remaining $11$ will be assigned to Group 1. The probability that person B is one of them is $\frac{3}{11}$. So the probability that A and B end up in different groups is $\frac{8}{11}$. There are more elaborate combinatorial arguments. Remark: At the request of ...

0

Here's a rough estimate which may give an idea of how to get a proper answer. Let $X_i(t)$ be the number of colors such that there are $i$ balls of that color. Let the level at time $t$, denoted by $L(t)$, be the largest $i$ such that $X_i(t)>0$. Suppose $L(t)=i$. We want the probability that $L(t+1)=i+1$. At time $t$, there are $i X_i(t)$ balls of ...

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$\int_0^{\infty}(x+1)e^{-x}dx=-(x+1)e^{-x}\Big|_0^{\infty}+\int_0^{\infty}e^{-x}dx=1+1=2$

1

Your answer is right. As for evaluating the integral, the limit at $\infty$ does exist, because $e^{-y}$ converges to zero as $y \rightarrow \infty.$ I think you should get an answer of $2$ after evaluating the integral, but you should check!

-5

The question doesn't really allow for a proper answer. After any number of coin tosses, you can expect one or more heads with some degree of confidence, that confidence being 1 minus the probability of flipping all tails.

3

Very interesting problem, and this is rather an write-up of a computation experiment to explore it. As it has been already pointed out in the comments, the dynamics of the urn can be described by a Markov chain on integer partitions of $n$. Let $\{m_1, \ldots, m_k \}$ be such a partition. Suppose $i$, $j$, such that $1 \leqslant i,j\leqslant k$, are the ...

1

Theoretically, this kind of question is impossible to answer, because there may be unknown factors. Suppose Team D has a weakness against Team A, so that it is more likely to lose against Team A than Team B, even though Team A will only beat Team B half of the time. This skews the probabilities, making an exact calculation impossible. But let's make the ...

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Team D is still twice as likely to win as A and C, even though B is now out of the running. A and C have equal odds still. Letting $P_{X}$ denoting the probably that team X wins, then we can set up the equation $P_{D}+P_{A}+P_{C}=1$. But again we already know $P_{D}=2P_{A}=2P_{C}$, so we can replace $P_{A}$ and $P_{C}$ in the first equation with ...

-1

Before the evening semi final between C and D, Probability of A winning is 40% (sum of probabilities of A and B), C winning is 20% (unchanged) and D winning is 40% (unchanged). Probability of D winning the evening semifinal should be 66.66%. ( It has double the chances as compared with the opponent C) Let us understand this situation with an example. let ...

2

As Jean-Claude Arbaut pointed out in a comment, $$\frac{n!}{n^{10000}{(n-10000)}!} = \frac{n}{n}\cdot\frac{n-1}{n}\cdots\frac{n-9999}{n}$$ If we make the approximations $n(n-9999) \approx (n-1)(n-9998) \approx \ldots \approx(n-4999.5)^2$, we get $$\frac{n!}{n^{10000}{(n-10000)}!} \approx \left(1-\frac{4999.5}{n}\right)^{10000} \approx 1 - ... 1 If you expand the factorial on the top and then cancel the factorial in the bottom, you have a numerator of$$(36^8)(36^8-1)(36^8-2)\cdots(36^8-9999)$$As 36^8 is much larger than 9999, this numerator is going to be only slightly smaller than (36^8)^{10000}. Hence we are subtracting off a number only minimally smaller than 1. This leaves us with a ... 1 Suppose that a certain set of balls has been drawn in the first round. We find the probability of no overlap in the second round. There are \binom{N}{k} equally likely ways to draw k balls in the second round. The number of no overlap ways is \binom{N-k}{k}. For the probability of no overlap, divide. Remarks: 1. If a\lt b, the binomial coefficient ... 0 Hint: What is the probability that no number is drawn twice in 2 consecutive rounds? 1 All these depend a lot on various circumstances that may or may not be the case in a particular application. For example, in (2) the Poisson model may be good if traffic is light and flowing freely, but not if it's a two-lane highway with few passing opportunities where traffic tends to bunch up behind the slower vehicles. Poisson would not be a good model ... 0 It follows from the very definition of a stopping time (page 155) that N \geq 1. Hence,$$-1_{\{N \geq 1\}} = -1.$$3 To make ends meet... You have been explained by several users that, looking at the toss process itself, one sees that the expectation E(X), that you know is E(X)=S, with$$S=\sum_{n\geqslant1}\frac{n}{2^n},$$solves the relation$$E(X)=1+\frac12E(X).$$It happens that one can also show directly that$$S=1+\frac12S,$$this relation following from a shift ... 7 Let X be the number of tosses, and let e=E(X). It is clear that e is finite. We might get a head on the first toss. This happens with probability \frac{1}{2}, and in that case X=1. Or else we might get a tail on the first toss. In that case, we have used up 1 toss, and we are "starting all over again." So in that case the expected number of ... 4 Your approach is perfectly fine. The probability of getting the first head in the nth trial is \frac{1}{2^n}, so we have$$ \mathbb{E}(x) = \sum_{ n \geq 1} \frac{n}{2^n}. $$This infinite sum can be calculated in the following way: first note that  \frac{1}{1-x} = \sum_{n \geq 0} x^n  for |x|<1. Differentiating both sides yields$$ ...

0

I now have: \begin{align} \mathbb{E}[\Phi (X,\mu^*)] &= \mathbb{E}[(\beta X_t) \mathbf{1}\{X_t<\mu^*\} + (\beta X_t+1-\beta)\mathbf{1}\{X_t\geq \mu^*\}]\\ &= \mathbb{E}[ (\beta X_t) \mathbf{1}\{X_t<\mu^*\} + (\beta X_t) \mathbf{1}\{X_t\geq \mu^*\} + (1-\beta) \mathbf{1}\{X_t\geq \mu^*\}]\\ &= \beta \mathbb{E}[X] + ...

1

You should probably refer to Casella and Berger, Exercises 2.34 and 2.35, as well as Section 2.6.1 after the exercises, where they discuss the uniqueness of moment sequences. There are several references mentioned in the text, specifically that of Romano and Siegel (1986), Counterexamples in Probability and Statistics. For your convenience, I have ...

0

The Theoretical probability for both strategies is the same. If in your experience, you find one strategy works better than the other, you are talking about Empirical probability. The dice roll is probably the classic example. The theoretical probability of a 6 showing on a roll is $\frac{1}{6}$. However, if you roll a dice 6 times and you get 2 sixes, you ...

0

Suppose four unprepared students sit for the test, say $W,X,Y,Z$. They decide that $Z$ will always guess $A$, $Y$ will always guess $B$, $X$ will always guess $C$, and $W$ will always guess $D$. On any given test, it is unlikely that all four students will get the same score. At random one of them, say $Y$, may get slightly more than 25%, while another, ...

0

Both will give an the same probability of securing marks. (Both will give you an expected marks of 25%.) The key thing to note is that each question is independent of another. This means that if the answer to the first question is B, then it does not make the answer to the second question more likely to be B (or any other choice, for that matter). Thus, ...

0

For the second question we can think of each pair of siblings as one "unit". Thus we have 5 units we need to place in 5 pairs of seats. Thus there are 5 ways to place the first "unit", 4 ways for the next, etc. However, there's two ways to arrange each unit, either sibling A first then sibling B or sibling B first then sibling A. So we need to consider both ...

1

Let's start with an answer for 1). There are 10! ways for all the 10 people to sit. That's because there are 10 seats for the first person to choose from, 9 remaining seats for the second to choose from, 8 for the third, etc. This is the denominator because it is the count of all possibilities. Since the seats are linear there are eight seats with a seat ...

0

1) The number of ways in which 10 people can sit is 10!. There are 9 ways for A and B to sit together, but you must also count the cases A-B and B-A. However, we don't care about the way the 8 people are sitting, so the number of ways for that is 8!, which means the result is $9*2*8!/10!$. 2) The number of possible outcomes is 10!; the number of ways for ...

0

Hints: 1) Count the number of successful outcomes, i.e. the number of ways A and B can sit to each other, then take into account that the remaining people can sit as they want, 2) There are 5! ways for pairs to sit together, 3) You can rotate the circle and the position remains the same

3

(Note: This is the solution to the problem in its original form.) Denote by $q(n)$ the probability that we hit $10$, given that the momentary sum is $n$ and we have not hit $10$ before. Then $$q(10)=1;\qquad q(n)=0\quad(11\leq n\leq14)\ .$$ Furthermore we have the following backwards recursion: $$q(n)=\sum_{k=1}^5 {1\over5} q(n+k)\qquad(n=9,8,7,\ldots)\ .$$ ...

2

Consider $X_n\equiv 1+\log n$ everywhere, and $\lambda=1$. Then every element belongs to LHS but RHS is empty... so this is a counterexample.

1

Let X be the discrete distribution which takes values 1 and 2 with equal probability. Then $E (X)=\frac32$ but $E (\frac1x) = \frac34$. (Almost any distribution you choose, discrete or continuous, will confirm that $E(\frac1X)\ne\frac1{E(X)}$. The underlying reason is that $\frac 1a + \frac1b \ne \frac1{a+b}$.)

0

Consider a discrete $X$ which takes value $-2$ or $2$, both with probability $\tfrac 12$. What is $1/X$ then? And what are $E$ of the two...?

1

For $N$ the expected value is the expected value until first seeing a number $\geq N$. For instance, for $N = 10$ and dice results $4,5,3,5,3,...$ the answer would be $3$, because $4+5 < 10,\text{ and } 4+5+3>10$, and for $5,6,3,2,1...$ it would be $2$. The bound on the expected value from above would be 10 (all 1's) and from above would be 2( you ...

6

It pays to generalize. Let's calculate the probability $p(n)$ that we ever reach $n$ for *any integer * $n$. Since we start at zero, we have $p(0)=1$, while $p(n)=0$ for $n<0$. For larger $n$, by conditioning on the previously taken value we get $$p(n)=\sum_{j=0}^5 p(n-j)/6,$$ and if you solve this recursive equation for $n=10$ you get ...

1

You should understand the difference between a probability density function and a cumulative distribution function. The cumulative distribution function, which in your case is $F(x)$, always gives the value for $P(X \leq x)$ So, $F(1)$ would give you $P(X\leq1)$ and $F(\frac{1}{2})$ would give you $P(X\leq\frac{1}{2})$. In order to find $P(\frac{1}{2} < ... 3 Counter-example 1: If$X \sim \text{Bin}(n,p)$, then$\mathrm{E}(X)=np$, but$\mathrm{E}(1/X)$is not even well defined:$k=0, \, 1/k=?$. Counter-example 2: If$X \sim \text{Bin(n,p)}$, then$\mathrm{E}(X+1)=np+1$, $$\mathrm{E}\left( ... 3 The star method: consider 22 balls and 3 separations (because you have 4 boxes). I denote * for the balls and \Big | for the separation. Then it's the number of permutation of:$$\left\{\underbrace{*\ *\ \cdots *\ }_{22\ balls}\Big|\hspace{0.5cm}\Big|\hspace{0.5cm} \Big|\hspace{0.5cm}\right\}$$There is 25! permutations but the permutation of the ... 1 How to use the stars and bars method? For x_i\ (i=1,2,3,4)\in\mathbb N, we have$$x_1+x_2+x_3+x_4=22\iff (x_1-1)+(x_2-1)+(x_3-1)+(x_4-1)=18.$$Here, note that x_i-1\ (i=1,2,3,4) are non-negative integers. Choosing 4-1=3 places (for bars) from 18+(4-1) places (for bars and stars) leads the answer is \binom{18+(4-1)}{4-1}=\binom{21}{3}=1330. ... 2 Yes, the Stars-and-Bars approach works great here, but you should know that there are two "versions". In both versions, we look for the number of distinct integer solutions to an equation such as yours. In the first version, every x_i must be a positive integer. In the second version, the restriction eases to include non-negative x_i. So, for example ... 0 I think your argument is correct. The book may use the Borel-Cantelli rather than subadditivity. That is ,there's a subsequence that \sum_{n_k} P(A_{n_k}^c)<\infty. Thus P(\limsup A_{n_k}^c)=0. Which means,$$P\left(\bigcup_{j=1}^\infty\bigcap_{k=j}^\infty A_{n_k}\right)=1.$$Let E_j=\bigcap_{k=j}^\infty A_{n_k} and we know that E_j\subset ... 1$$\{X\leq 1\}=\{X<\frac{1}{2}\}\cup\{\frac{1}{2}\leq X\leq 1\}$$and these sets are disjoint so that:$$P\{X\leq 1\}=P\{X<\frac{1}{2}\}+P\{\frac{1}{2}\leq X\leq 1\}$$or equivalently:$$P\{\frac{1}{2}\leq X\leq 1\}=P\{X\leq 1\}-P\{X<\frac{1}{2}\}$$0 We cannot substitute 2 into F(x) because F(x) is by definition equal to \Pr\{X\le x\} and we are looking for the probability \Pr\{X\ge2\}. We have that$$ \Pr\{X\ge2\}=1-\Pr\{X<2\} $$and$$ \Pr\{X<2\}=\Pr\{X\le2\} $$since the random variable X is continuous. So the answer is$$ 1-F(2).$$2 Prescribe measure$\nu$by$\nu(A)=0$if$\mu(A)=0$and$\nu(A)=+\infty$otherwise. This for instance where$\mu$is the Lebesguemeasure on Borelsets in$\mathbb R$. Then$\nu\ll\mu$,$\mu$is$\sigma$-finite but$\nu\$ is not.

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