Tag Info

3

There are 3 $A$s and $8$ possible spots for them, so the probability that they will be in the correct spots is $\frac{1}{8\choose3}$. Given that this has happened, there are $5$ spots for the two $B$'s so the probability that they are in the correct spot is $\frac{1}{5\choose2}$. Given that both of these have happened, the $R$s must be in the correct spots, ...

2

"The probability of the second person having a birthday on a day different than the first is 1/364." Is it? Regarding the rest of your question, you should consider permutations not combinations.

2

For question 1, you cans use $E[\frac{X-\mu}{\sigma}]=E[\frac{X}{\sigma}]-\frac{\mu}{\sigma}=\frac{\mu}{\sigma}-\frac{\mu}{\sigma}=0$. For the variance, use the variance properties: $Var(\frac{X-\mu}{\sigma})=Var(\frac{X}{\sigma}-\frac{\mu}{\sigma})=Var(\frac{X}{\sigma})=\frac{1}{\sigma^2}Var(X)=\frac{\sigma^2}{\sigma^2}=1$. Question 2 can be answered in a ...

2

There is no need to use PDFs. Since you're only allowed to use properties of expectation, then $$E[Z] = E\left[\frac{X-\mu}{\sigma}\right] = \frac{1}{\sigma}\left[E[X] - \mu\right] = \frac{1}{\sigma}[\mu-\mu] = 0.$$ Then for the variance \begin{align*} \text{Var}[Z^2] &= E[Z^2]-E^2[Z] \\ &= E\left[\left(\frac{X-\mu}{\sigma}\right)^2\right] - ...

2

I hope this isn't just semantics and that I'm reading you're question correctly, but the expression you write is less of a "fundamental property" of $\mathbb{E}[X\vert Y]$ than it is a definition. Intuitively, $\mathbb{E}[X\vert Y]$ denotes the expectation of X conditional on the "$\sigma$-algebra" generated by the random variable $Y$ (which we denote ...

1

The first player has a 1/2 + 1/8 + 1/32 + 1/128 + .... = 2/3 chance of winning and the second player has a 1/4/ + 1/16 + 1/64 + ... = 1/3 chance of winning. This is kind of interesting in that it means if the two players played a game where they flip a coin until it was heads; the first player wins if the total number of flips is odd and the second player ...

1

The answer is the same as the first part: the bigger deck, because no new information is revealed to you when a bunch of the cards are thrown out of the big deck. Perhaps it is simpler to see if you think about what is happening when you randomly choose two cards. It's kind of like you are throwing out 50 of the original 52. It doesn't make a difference if ...

1

I think this is a good question, as it points to an interesting conceptual problem about what a parameter is. In general, if we talk about the probability density function (PDF) of a particular distribution, we usually refer to one established of infinitely many possible forms of PDFs that describe the distribution equally well. Let me explain this by the ...

1

The Chebyshev inequality is $$\mathbb{P}(|x - \mu| \geq a) \leq \frac{\sigma^2}{a^2}$$ .Substituting $$a = k\sigma$$gives the answer.

1

Thanks Andre, I found it. For any event A, let IA be the indicator random variable of A, i.e. IA equals 1 if A occurs and 0 otherwise. Then

1

Yes. It will converge a.s. to the zero (constant) random variable. This is a simple application of the Borel-Cantelli lemma. Let $A_n$ be the event that $X_n=1$. Then $\mathrm{Pr}(A_n) = \frac{1}{2n^2}$, which is summable. Hence by the Borel-Cantelli lemma, with probability one, only finitely many of the $A_n$ will occur. Note that independence of $X_n$ is ...

1

a) We are choosing numbers one at a time. Whatever collection of numbers we chose, by symmetry the probability the first chosen number was biggest is $\frac{1}{4}$. We can also make a calculation based on your sample space $\Omega$. As you pointed out, this sample space has $(10)(9)(8)(7)$ outcomes. We now count the favourables. There are $\binom{10}{4}$ ...

1

For the sake of the other users to understand the solution to this question, I'll redo the entire problem. As you stated in your question, you identified that this problem requires Bayes Theorem and conditional probability. Let $A$ be the event that the chosen employee is a woman, and $B$ be the event that the chosen woman came from the store with $12$ ...

1

"The probability of the second person having a birthday on a day different than the first is 1/364." This is false. Consider the case where the first two people shared the same birthday. You may also note that $Q(n) = \frac{1}{P(n)*365^n}$ if you wish to deduce $Q(n)$ from $P(n)$.

1

Something is very, very, very wrong here. The probability of the second person have a birthday on a day different than the first is 364/365 = 1 - 1/365. That lecturer needs to learn just a little bit to use intuition. The lecturer suggests the probability that three random people have their birthdays on different days is (1/365) * (1/364) * (1/363). That's ...

1

I saw that your last comments suggested that you still are having trouble with these so let's talk about it. What's the general theory? So let's say that you have some $X, Y$ with joint probability density $$j(x,\, y) = \lim_{dx,dy\to0}\frac1{dx~dy}\text{Pr}\big[X \in (x,\, x+dx) ~\land~ Y \in (y,\, y+dy)\big]$$For independent random variables with PDFs ...

1

By the Inclusion-Exclusion principle, it is: 1-4*Pr(A is void in spades)+6*Pr(A,B are void)-4*Pr(A,B,C are void) $A$ is void in spades with probability ${39\choose13}/{52\choose13}$. $A,B$ are void in spades with probability ${39\choose 26}/{52\choose26}$. Step 1: 1-Pr(A is void)-Pr(B is void)-Pr(C is void)-Pr(D is void). Step 2: We double-removed the ...

1

Count all $8$-long sequences with elements from $\{2,3,4\}$ and divide that number by the number of all $8$-long sequences with elements from $\{1,2,3,4\}$. $$P=\frac{3^8}{4^8}$$ The nominator is the number of outcomes that satisfy your needs, and the denominator is the total number of possible outcomes. These outcomes are equally likely. Does this make ...

1

Here is an outline/hint: What is the relationship between the radius and the area? $$A = \pi R^2.$$ They tell us that $R\sim\text{unif}(a,b)$. Now, I need to find the distribution of $A$ using that relationship. What do I do?

1

You have written your conditional density incorrectly. It should be $f_{X\mid Y}(x,y) = \frac1y \mathbb 1_{[0,y]}$, i.e. it is $\frac1y$ on $[0,y]$ and $0$ outside. Then, \begin{align} E[X] &= E[E[X\mid Y]]\\ &=E\left[\int_0^\infty xf_{X\mid Y}(x,y) dx\right]\\ &=\int_0^1\left[\int_0^y \frac{x}{y} dx\right]dy\\ &=\int_0^1\frac y2dy\\ ...

1

If you have two distributions $X\sim\mathcal U(a;b), Y\sim\mathcal U(c;d)$ then on a plot of $y$ versus $x$, the density function $f_{X-Y}(u)$ will be proportionate to the measure of the length of the segment of that passes through the joint support rectangle $(a..b){\times}(c..d)$ along the line $y=x-u$ . Clearly, as you said, these segments will be ...

Only top voted, non community-wiki answers of a minimum length are eligible