# Tag Info

0

Here's an attempt at a reasonable mathematical model for this. We'll suppose you're on a circular road with $N$ parking spots (numbered $0$ to $N-1$) and $N$ corresponding waiting positions (also $0$ to $N-1$) for your car. At each waiting position $x$ , you can see $m$ parking spots ahead (positions $x$ to $x+m-1 \mod N$). Yours is one of $p$ cars ...

1

Everything is good up to your inequality. You have correctly shown that $$P(N(s+t)=k | N(s)=j, \{N(u), 0<u<s\}) = P(N(s+t)-N(s)=k-j).$$ Now start with only $P(N(s+t)=k | N(s)=j)$, without the $\{N(u), 0<u<s\}$ and show that it is also equal to $P(N(s+t)-N(s)=k-j)$. Then you are done!

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Hint. There are more yellow apples than red apples if you either pick two yellow apples and one red apple, or three yellow apples and no red apples. What is the probability of doing each of those things? Add them up and you'll get your answer.

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Both $(2)$ and $(3)$ are correct. You might use $(2)$ if you know something about conditional probabilities given $C$. All probabilities are conditional. In $(1)$, the events $A$ and $B$ are subsets of some larger probability space (or you might know it as a "sample space"), which let us call $\Omega$. Thus you can write $(1)$ as $$P(A \mid B,\Omega) = ... 1 Use the Law of the Unconscious Statistician, and find \iint_S u\sin(vt)\,du\,dv on the square S. 0 The simplest way to do (a) is to find the probability of answering 0 or only 1 correctly, then subtracting that from 1. With four possible answer to each problem, the probability of answering one correctly by choosing at random is 1/4 and the probability of answering incorrectly is 3/4. So the probability of answering all 8 questions incorrectly (0 ... 0 HINT Let Y have exponential distribution. What is the pdf f_Y(y) and the cdf F_Y(y)? The hazard rate is defined by$$ h(y) = \frac{f_Y(y)}{1-F_Y(y)}. $$Plug both into this definition and simplify. 4 E.g. let \vec{Y} be a random vector with normal distribution on \mathbb R^n with \mathbb E\vec{Y}=\vec{0} and the identity matrix as covariance matrix. Then \vec{X}:=\vec{Y}/||\vec{Y}|| is a uniformly distributed unit-vector. 1 In principle you can do this using a (rather large) Markov chain. You have 64^2 states of the form (i,j,k,l) indicating that the ants are at (i,j) and (k,l). If A is the transition matrix then I think you'll need to look at \frac{d}{dt}(I-tA)^{-1} at t=0. But A is a 4096\times4096 (sparse) matrix, so this is a challenge. It would make a ... 0 The sum of the squared moments is \begin{eqnarray*} \sum_{k=0}^\infty \mathbb{E}(X^k)^2&=& \sum_{k=0}^\infty \left(\int_0^1 x^k\,\mu(dx)\right)^2\\[5pt] &=& \sum_{k=0}^\infty \int_0^1\int_0^1 x^k y^k\,\mu(dx)\,\mu(dy)\\[5pt] &=& \int_0^1\int_0^1 {1\over 1-xy}\,\mu(dx)\,\mu(dy). \end{eqnarray*} This will be finite if \mu doesn't put ... 0 I can give a proof with one additional assumption on f. First consider Holder's inequality for an arbitrary p \in [1,\infty], with Holder conjugate q:$$\left ( \int_0^1 x^n f(x) dx \right )^2 \leq \| x^n \|^2_p \| f \|^2_q \leq C \left ( \int_0^1 x^{np} dx \right )^{2/p} = C (np+1)^{-2/p}.$$We want this exponent to be strictly less than -1, so we ... 0 Notice that:$$\max(1, \min(Y + a,0)) = 1$$for every a and Y. Indeed: If Y+a \geq 0, then \min(Y + a,0) = 0 and \max(1, 0) = 1. If Y+a < 0, then \min(Y + a,0) = Y + a and \max(1, Y + a) = 1. Then you have:$$\Pr[ X + \max(1, \min(Y + a,0)) \leq b \,|\, Y > b - c ] = \\=\Pr[ X + 1 \leq b \,|\, Y > b - c ] = \\=\Pr[ X \leq b-1 ...

0

For you Probability and Random processes by Grimmet seems to be a good starting place. Along with that you can read these lecture notes by Bruce Hajek. Once you've got reasonable hold on these books you can go for studying some measure theory since without it, it is quite impossible to learn advance graduate level probability which you might be requiring ...

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The joint distribution of inter-arrival times is uniform on $\{x_1+\dots + x_n<t\}$ conditionally on the fact that $N(t) = n$. Unconditionally, they are independent and have exponential distribution. There is no contradiction; you can try to verify this fact yourself, it's not that hard. Moreover, as @Did wrote, while the joint (conditional) distribution ...

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Hint: $N$ is distributed as $P_1 + \dots + P_A$, where $P_k$ are iid $\mathrm{Poisson}(\lambda)$. So by the law of large numbers, $N/A \to \lambda$, $n\to\infty$, in probability.

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The number is close enough to $0$ that you can make some approximations: given any pair, the probability they coincide is $1/(18 \times 10^{18})$ and there are about $12.5 \times 10^{12}$ possible pairs so the expected number of coincidences is about $6.944 \times 10^{-7}$ and since this is small it is also the approximate probability of any coincidence. ...

0

Two things that may help: $v(x) \geq 0\mbox{ for all } \{x:0\leq x \leq 6\}$ (Probability distributions must be non-negative on the domain) $\int_0^6v(x) = 1$ With this in mind I would first look at where $t \sin(t) = 0$. What value of $n$ would make $6\sin(n6) = 0$ while keeping the distribution non-negative over the domain? (Technically there are two ...

2

A) For the probability $P$ has more elements than $Q$, recall from the answer to the earlier problem about the probability $P$ and $Q$ are of equal size is $a$, where $a=\frac{\binom{2n}{n}}{2^{2n}}$. So the probability $P$ and $Q$ are unequal is $1-a$, and therefore by symmetry the probability $P$ has more than $Q$ is $\frac{1-a}{2}$. B) This yields to an ...

1

Start from simple case $G(1)$. We know there is only one strategy and the expected value $E(1)=\frac{1+4+9+16+25+36}{6}=\frac{91}{6}$ For $N=2$, there are essentially $6$ possible strategies: Switch when roll is 1, $E_1=\frac{\frac{91}{6}+4+9+16+25+36}{6}$ Switch when roll less than 3, $E_2=\frac{\frac{91}{6}\cdot2+9+16+25+36}{6}$ Switch when roll less than ...

2

The numerator is not correct. Note that the 9-digit numbers cannot use digits more than once. Hints: The number in the center (the 5th position) must be odd. (Why?) After choosing the number in the center, the four [unordered] pairs of digits are fixed. For example, if $1$ is in the center, then you must have $(2,3)$, $(4,5)$, $(6,7)$, and $(8,9)$ as your ...

2

I assume we pick $P$ and $Q$ uniformly and independently from the power set of $A$. Since there are $n\choose k$ subsets of size $k$ the answer to the first question is $$\sum_{k=0}^nP(|P|=k)P(|Q|=k)=2^{-2n}\sum_{k=0}^n{n\choose k}^2$$ (the expression can still be simplified, cf. André's explanation) For the second question, note that each element of $A$ ...

3

First problem; It is convenient to imagine that the choices are made from two distinct $n$-element sets, $A$ and $B$, say boys and girls. There are $2^n$ equally likely ways to choose a subset of $A$, and $2^n$ equally likely ways to choose a subset of $B$, for a total of $2^{2n}$. There are $\binom{n}{k}$ ways to choose a $k$-subset of $A$, and ...

0

No, the supremum is over $n$. For each $\omega \in \Omega$, $$(\sup_{n \in \mathbb{N}} X_n)(\omega) := \sup\{X_n(\omega):n \in \mathbb{N}\}$$ Note that $\sup_n X_n$ is still a random variable! Similarly, $\limsup X_n$ is a random variable. For each $\omega \in \Omega$, $$(\limsup_{n \to \infty} X_n)(\omega) := \limsup_{n \to \infty} X_n(\omega).$$ That is, ...

0

Comments: (1) I suppose you may already know that this is closely related to the famous birthday problem. (2) If $n$ is moderately large compared with very large $k,$ then $Y = n - X$ (the number of redundancies) may be approximately Poisson. (3) If $k = 20$ and $n = 10,$ then your formula has $E(X) = 8.0253$. A simulation of a million such experiments ...

1

"What is the probability that the President is a computer science major given that exactly two officers are computer science majors?" You have 5 officers and you say exactly 2 are CS majors. That's enough information to answer the question. You can ignore all the stuff leading up to that. What is the probability that the a particular officer is one of ...

1

(a) is correct. The favoured space are ways to select two computer majors and thee non-computer majors in the other position. The total space are ways to select any five students for the position. $$P(A) = \frac{\binom{8}{2}\binom{22}{3}}{\binom{30}{5}}$$ For (b) When given 2 computer majors and 3 other students, what is the probability that the one who ...

1

You have not given anywhere near enough information for me to give you a responsible answer. So I will make some very speculative assumptions, and give you an answer based on those assumptions. If you are not satisfied with the answer, maybe that will prompt you to provide more insight into what your data are like and what you really want to do. I assume ...

0

Consider the bulbs to be distinct individuals, so that each selection of two individuals are equally probable events. How many ways can you so select two bulbs from the failures? How many ways can you so select two bulbs from all the bulbs? Go forth and divide!

2

Almost.   You have the probability that two cards come from a particular suit.   You also have to consider that there are four suits from which those cards can be selected. Further, we typically use combination rather than permutation for this, because order of selection in a card hand is not important.   (Hands are "heaps" rather than ...

0

For this particular question it does not matter whether you do permutations or combinations, as long as you are consistent about using the same counting method in the numerator and the denominator. Your method with permutations is correct (up to the point where you forgot to account for the other three suits), exactly as you showed it. If you used ...

0

You have calculated the probability of drawing two cards from one particular suit, but there are 4 suits to consider. Then there are actually $4*_{13}P_2$ ways to select two cards of the same suit.

1

As has been pointed out, the question makes little sense; your $X$ is a uniform random variable by definition, and that has nothing to do with those $\omega_j$ things. You meant to ask this: Saying $d_j$ are iid random variables, with $P(d_j=0)=P(d_j=1)=1/2$. Define $$X=0.d_1d_2\dots=\sum_{j=1}^\infty2^{-j}d_j.$$Is $X$ uniformy distributed on $[0,1]$? ...

0

You want to chose one type of tile for the pair ($34$ possibilities) and ${4 \choose 2}=6$ ways of choosing two of that type. Although you pick in groups of four, you can imagine this as picking singly so they can have been picked in any position for your hand so $14\times 13=182$ pairs of positions You then want to choose $12$ types from the remaining ...

2

Consider a Poisson process of rate $\lambda$. Independently, each occurrence of the Poisson process is "special" with probability $q = 1/(p+1)$. Your $T$ is the waiting time until the first special occurrence. You can also consider this from a different point of view: the special and the non-special occurrences form independent Poisson processes with ...

0

You can use MGFs to obtain the distribution of the conditional sum $X \mid N$ and show that this is gamma distributed with rate $\lambda$ and shape $N$. Then we would compute the unconditional/marginal distribution $X$ by summing over all $N = 0, 1, 2, \ldots$, weighted by the probability $\Pr[N = n]$.

3

If if doesn't have to be exact, the normal approximation will be easy to use. If an individual bit is chosen to be flipped, it has $\frac nm$ chance of starting as a $1$ and $\frac {m-n}m$ chance of starting as a $0$. It has $\frac n{2m}$ chance of changing from $1$ to $0, \frac {m-n}{2m}$ chance of changing from $0$ to $1$ and $\frac 12$ chance of not ...

1

Use $E(Y)=E(E(Y|X))=E(E(Fs(X)))=E(1/X)$ $Var(Y)=E(Var(Y|X))+Var(E(Y|X))=E((1-X)/X^2)+Var(1/X)$. Where $Var(Z)=E(Z^2)-E(Z)^2$ and $E(g(X))=\int g(x)f_X(x)dx$

1

First, you have $$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$$ Similarly, $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$ So the product is just $$P(A \mid B)P(B \mid A) = \frac{(P(A \cap B))^2}{P(A)P(B)} \geq 0.$$ Observe that $$P(A \mid B)P(B \mid A) \leq 1,$$ as we have $$A \cap B \subset A \Longrightarrow P(A \cap B) \leq P(A)$$ and $$A \cap B \subset B ... 0 If we assume the missiles to represent (independent) Bernoulli trials, the binomial distribution is applicable (since the failure rate 0.005 is so low, we could approximate this with the Poisson distribution for large n). The probability of k missiles out of n penetrating the defence system is:$$p(k,n)=\binom{n}{k}(0.005)^k(0.995)^{n-k}$$For ... 1 One example: the stable law with \alpha\in(0,2) has no moments of integer order greater than 1 (for \alpha\in(0,1] even the expectation does not exist). So in order to measure the "dispersion" one should look at the moments of lower order. Say, for \alpha\in(1,2) this would be E[|\xi-\mu|^p] with p\in(0,\alpha). 1 It is hard to understand why don't you use Alex R.'s hint, so let me write it down. By Levy's theorem, L_t has the same distribution as M_t = \sup_{s\in[0,t]} B_t. So by the law of the iterated logarithm,$$ \limsup_{t\to 0+} \frac{L_t}{\sqrt{2t\log \log \frac1t}} = \limsup_{t\to 0+} \frac{M_t}{\sqrt{2t\log \log \frac1t}}\ge \limsup_{t\to 0+} ...

1

You can think of $p(y|x,\theta)$ as the distribution of $y$ when $x$ is considered as a constant (i.e. it's not a random variable). $(2)$ means that $\epsilon\sim\mathcal{N}(0,\sigma^2)$. This implies (using the properties of gaussian distributions) that, $\theta^Tx$ being considered constant, $\theta^T x + \varepsilon\sim\mathcal{N}(\theta^Tx,\sigma^2)$, ...

2

There is no finite number of trials such that you are sure that the proportion of tails (or heads) is exactly 50/50. For any finite number of trials, the proportion of tails (or heads) is always a random variable. It can, by chance, be equal to 0.5 (for example, you try twice and you obtain one of each result), but there will always be incertainty. Even if ...

0

1a: $0.85 \times 0.87$ 1b: $1 - (0.15 \times 0.13)$

0

I know that $P(A∩B)=P(A)+P(B)−P(A∪B)$ . If I take 1/2 of the whole right hand side I can say $P(A∩B)≤P(A)/2+P(B)/2−P(A∪B)/2$ No. Let's put some number on that. You're saying that if $0.2 = 0.4 + 0.4 - 0.6$ then $0.2\leq 0.2 + 0.2 - 0.3 = 0.1$

1

The probability of rolling a 2 or 6 on a fair die is $1/3$ and the probability of rolling an odd number (so 1, 3 or 5) is $1/2$. Since these two events are independent, $P(A) = 1/2 \cdot 1/3 = 1/6$. In general with problems like these, you can sometimes split up the event $A$ into more manageable parts (divide and conquer) that are independent and then ...

2

To convert something to be "1 in N", you simply take the reciprocal. In this case, you have it correct: $$\frac{1}{0.00000007151123842} = 13,983,816$$ So you get an event that happens "1 in 13,983,816" tries. Odds are a bit different in that they usually give the ratio between two events happening. For example 1:1 (pronounced "1 to 1") odds means a 50% ...

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You can directly apply the special case of Chernoff bounds to this problem. A loose upper bound can be derieved from Chebyshev's inequality as well.

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There are 3 ways to chose the 1st and 5th digits, 4 ways to choose the 2nd and 4th digits and 3 ways to choose the 3rd, 6th and 7th digits. Thus there are $3^2\cdot 4^2 \cdot 3^3 = 3^5\cdot 4^2 = 3888$ different cards with the special property you described. In total there are $10^7$ possible card numbers, and assuming each number is equally likely the ...

0

First part: Let $a,b,c$ represent $3$ consecutive days. Since we are in state $1$, that means we have the sequence $(a,b) = \text{(no rain, rain)}$. In order to jump onto state $0$, there must hold $(b,c) = \text{(rain, rain)}$. Then we have the sequence $(a,b,c) = \text{(no rain, rain, rain)}$. According to the assumptions, starting from $(a,b)$ we can ...

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