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0

Placing $36$ cards in a row can be done on $36!$ ways. If we add the condition that the first $4$ cards are aces then it can be done on $4!32!$ ways. If we add the weaker condition that the aces must be adjacent then it can be done on $33.4!32!$ ways. So there is a probability of: $$\frac{33.4!32!}{36!}=\frac1{1785}$$ that it happens.


0

Your notation is mildly confusing for a reason I will explain further below. You have $X_t = \sqrt{t} Z$ where $Z\sim\mathrm{N}(0,1)$ and $0<s<t$, so that $X_t \sim \mathrm{N}(0,t)$. It follows that $X_t-X_s = \sqrt{t}Z-\sqrt{s}Z$ $=(\sqrt t - \sqrt s)Z$, so the distribution of this random variable is $\mathrm{N}(0,(\sqrt t - \sqrt s)^2) = ...


1

The hint would suggest looking at $E[X^2] = E[X\times X] \le E[cX] = cE[X]$ since $0 \le X \le c$. Your next stage would be to look at $\text{Var}(X)=E[X^2] - E[X]^2 \le cE[X] - E[X]^2 = c^2\alpha - c^2\alpha^2 = c^2[\alpha(1-\alpha)]$ using the suggestion involving $\alpha$. Now take the derivative with respect to $\alpha$ of $c^2[\alpha(1-\alpha)]$ to ...


0

Your mistake is the number of ways to place $4$ cards in a row. I assume this is $4$ cards out of the deck of $36$. There are $36$ choices for the first card, then $35$ for the second (since the first is no longer available), then $34$ for the third, then $33$ for the fourth. Thus the number of ways to choose $4$ cards out of the deck of $36$ and place them ...


1

Suppose it is not true. Then $P(A_1\mid B)<P(A_1)$ and $P(A_i\mid B)\leq P(A_i)$ for $i=2,\dots,k$ leading to: $$1=P(\Omega\mid B)=\sum_{i=1}^kP(A_i\mid B)<\sum_{i=1}^kP(A_i)=P(\Omega)=1$$ A contradiction is found.


0

Something feels off here, why would it be $\sqrt{N}$? Are you sure you've stated the problem correctly, where each column has only one color? Notice that this process is reversible: you can start with 0 balls in both bins and add balls with equal chance on each step, and when a column fills up, you must put balls in the non-full one. To see this, imagine ...


0

Let $Z_{n,j}$ be i.i.d. with common distribution $$ \mathbb P(Z_{n,j}=0)=\frac15 = 1- \mathbb P(Z_{n,j}=3)$$ and $X_{n,j}$ be i.i.d. with common $\mathrm{Ber}\left(\frac12\right)$ distribution for positive integers $n$, $j$, and $k=1,2,3$. Assume that we start with one male and one female rabbit. The population in generation $n$ is given by $Z_0=1$ and ...


1

Approach to solution: First, determine the CDF of the needed exponential distributions $$ F(x) \equiv P(X_i < x), \qquad i = 1, 2 $$ You will need to figure out the actual formula for $F(x)$; the above is only its definition. Now, determine $G(x) \equiv P(T < x)$ for any $x$, the CDF of the maximum of $X_1$ and $X_2$. Note that $G(x) = P(X_1, X_2 ...


1

Note that $$ -xe^{-x^2/2}\,\mathrm{d}x=\frac{\mathrm{d}}{\mathrm{d}x}e^{-x^2/2} $$ Since $\Phi(0)=\frac12$, we can integrate by parts to get $$ \begin{align} \int_0^\infty x\Phi(x)\phi(x)\,\mathrm{d}x &=\frac1{\sqrt{2\pi}}\int_0^\infty x\Phi(x)e^{-x^2/2}\,\mathrm{d}x\\ &=-\frac1{\sqrt{2\pi}}\int_0^\infty \Phi(x)\,\mathrm{d}e^{-x^2/2}\\ ...


1

Have you tried integration by parts together with the knowledge that the antiderivative of $x\phi(x)$ is $-\phi(x)$? The answer is readily computable via this approach.


2

Hint: $T=\max\{X_1,X_2\}$. So it would be wise to first deduce the distribution of $T$ and then obtain the expectation of $T$. Notice that $X_1,X_2$ are independent with exponential parameter $\dfrac{1}{60}$. Can you take it from here?


0

1)If only two characters "$x$" and "$1$" are allowed, then we can have exactly $2^8$ strings. 2)There are in total $36^8$ strings of length $8$. The number of strings with exactly one "$y$" is $8\cdot 35^7$, that is there are $8$ possible places for the "$y$", and in the rest we can use $35$ characters. Hence, the required probability is $\frac{36^8}{8\cdot ...


1

Let $n\ge 3$. Number the seats from $1$ to $n$. Define indicator random variables $X_1,\dots,X_n$ by $X_i=1$ if the person in seat $i$ is shorter than her two neighbours, and let $X_i=0$ otherwise. We want to find $E(X_1+\cdots+X_n)$. By the linearity of expectation this is $E(X_1)+\cdots+E(X_n)$. We have $\Pr(X_i=1)=\frac{1}{3}$, so ...


2

Pick a person. Among that person and the two neighbours, exactly one is the most vertically challenged. The probability that it is the initially picked person is $\frac13$ simply by symmetry. Hence the expected number is $\frac n3$ (assuming $n\ge 3$).


0

The distribution of $X+Y$ is a binomial distribution, as expected: $$ q_0 = q_3 = 1/8 \\ q_1 = q_2 = 3/8 $$ where $q_i = P(X+Y = i)$. This can be reasoned out as follows: You can obtain $0$ only if $X = Y = 0$, so the probability is $(1/4)(1/2) = 1/8$. You can obtain $3$ only if $X = 2, Y = 1$, so again the probability is $(1/4)(1/2) = 1/8$. Finally, you ...


0

I suppose, except measure theoretical discusses, that you know the probability function is countably additive, i.e for disjoint family of events $\{A_n\}_{n\in\mathbb{N}}$, we have $P(\bigcup A_n)=\sum_n P(A_n)$. In general case, suppose you have the nested events $\{E_i\}$, i.e $E_n\subset E_{n+1}$, then by defining $E_0=\emptyset \wedge A_n:=E_n\setminus ...


0

If $Y\sim U(a,b)$, then $Y$ has density $$f_Y(y)=\frac1{b-a}\mathbb I_{(a,b)}(y),$$ where $\mathbb I$ denotes the indicator function. Hence $-Y$ has density $$f_{-Y}(y) = f_Y(-y) = \frac1{b-a}\mathbb I_{(a,b)}(-y) = \frac1{b-a}\mathbb I_{(-b,-a)}(y).$$ It follows that $-Y\sim U(-b,-a)$. From here, you would compute the convolution $$(f_X\star f_{-Y})(t) = ...


3

The argument goes as follows: \begin{align} \textbf{Pr} \! \left( \bigcap_{i = 1}^{n} (\Omega \setminus A_{i}) \right) & = \textbf{Pr} \! \left( \Omega \setminus \bigcup_{i = 1}^{n} A_{i} \right) \qquad (\text{By De Morgan’s Laws.}) \\ & = 1 - \textbf{Pr} \! \left( \bigcup_{i = 1}^{n} A_{i} \right) \\ & = \prod_{i = 1}^{n} [1 - ...


1

$X-Y = X + (-Y)$. Find the probability density of the random variable $-Y$, then find the convolution.


2

Is it good answer? $$X=\{0,1,2\}, Y=\{0,1\}$$ $$P(Y=1)=1/2, P(Y=0)=1/2, P(X=0)=1/4, P(X=1)=1/2, P(X=2)=1/4$$ $$P(X+Y=0)=1/8, P(X+Y=1)=3/8, P(X+Y=2)=3/8, P(X+Y=3)=1/8$$ $$P(X-Y=-1)=1/8, P(X-Y=0)=3/8, P(X-Y=1)=3/8, P(X-Y=2)=1/8$$ $P(XY=0)=5/8, P(XY=1)=2/8, P(XY=2)=1/8$


0

Your sets are nested so the union is just the last set in the union for the finite unions, so the equality is justified.


0

Is the standard deviation of $600$ correct? If so, here are some hints: a) $z=\frac{630-650}{600}$. You need to find the area to the left of this. b) $z=\frac{630-650}{\frac{600}{\sqrt{70}}}$. Again, area to the left.


4

I will mark the set of grades by the three people as $\{a, b, c\}$, (where order doesn't matter). Note that, if $x, y, z$ are distinct, then: \begin{align} \Pr[\{x, x, x\}] &= 1 / 5^3 \\ \Pr[\{x, x, y\}] &= 3 / 5^3 \\ \Pr[\{x, y, z\}] &= 6 / 5^3 \end{align} Final Grade = 1 Requires $\{1, 1, 1\}$ or $\{1, 1, a\}$ for some $a \geq 2$. Ie, ...


2

In general, the claim does not hold true as the following counterexample shows: Consider $((0,1],\mathcal{B}(0,1])$ endowed with the Lebesgue measure. For $a_n := \frac{1}{n^2}$ we define $$g_n(\omega) := \begin{cases} \frac{1}{n}, & \omega \notin [a_n,a_{n-1}), \\ \frac{1}{\omega}, & \omega \in [a_n,a_{n-1}), \end{cases} \qquad \omega \in (0,1].$$ ...


1

Counterexample: Take $\Omega=[0,\infty)$ and consider the sequence of characteristic functions: $1_I, 1_{I/2+1}, 1_{I/2+1+1/2}, 1_{I/3+2}, 1_{I/3+2+1/3}, 1_{I/3+2+2/3}...$ Where $1_A$ is the characteristic function of the set $A$ and I is the unit interval.


0

By looking at your formula for the even case (in its infinite sum form), note that $$(1-p)P(E) = p(O) - p$$ As the odd and even case sums to 1, $$1= P(E) + p + (1-p)P(E)$$ or $$P(E) = \frac{1-p}{2-p}$$


1

Define $(1-p)=:r$, then $P(X=2n) =pr^{2n-1}$. We are interested in the probability, that $X$ is even, therfore, we calculate \begin{align} P_{even}&= \sum_{n=1}^\infty P(X=2n) \\ &= \sum_{n=1}^\infty pr^{2n-1}\\ &= p\sum_{n=1}^\infty r^{2n-1}\\ &= \frac{p}{r}\sum_{n=1}^\infty r^{2n}\\ &= \frac{p}{r}\sum_{n=1}^\infty (r^2)^{n}\\ ...


0

This is equal to (12C3 * 3C1 * 2^9)/ 3^12 which is inturn equal to 55/3 * (2/3)^11 hence option (2).


1

I have a simpler proof. I hope this is ok. Let $Y = X - a$ be a random variable. Now note that due to symmetricity $Y$ and $-Y$ have the same distribution. That implies $$E[Y^3] = E[(-Y)^3]$$ This implies $E[Y^3] = 0$.


0

Apologies for the long post but this material can be confusing so I tried to be clear by explaining every step. Probability that the student knows the answer $= P(KC) = p$ but not $P(C|K)$. It's not. The probability that the student knows the answer is $P(K)$, not $P(KC)$. The probability of the joint events student knows the answer and student answers ...


1

Imagine that the test consists of $N$ questions, each with the same parameter $p$ of the student knowing the right answer; and assume that knowing the right answer on any question is independent of knowing the right answer on any other question. In this scenario, each question will fall into one of three categories: (A) questions where the student knew ...


0

The way I work through these is by thinking "the probability of B GIVEN A is equal to the probability of A AND B divided by the probability of just A. So in this case, the probability that the student knew the answer given he answered it right is the probability he knew it and got it right divided by the probability he answered it right. So probability ...


0

In my class my teacher states that: Let $X$ be a random variable with $S_X=\{0,1\}$. $X$ follows a Bernoulli distribution if $P(X=x)=p^x(1−p)^{1−x}$ for some $p$ in $(0,1)$. Doubtful. Your teacher should have stated: Let $X$ be a random variable with support, $S_X=\{0, 1\}$.   Then $X$ has a Bernoulli distribution of $\mathsf ...


0

There are two distributions called Geometric. 1. The distribution of Bernoulli trials until a failure. ( This is sometimes called the Shifted Geometric Distribution. ) Assuming: $Z=X+Y$ and $\mathsf P(X=k)=\mathsf P(Y=k)=(1-p)p^{k-1}, k\in \{1,2...\}$ then $\begin{align} \mathsf P(Z=z) & = \sum_{k=1}^{z-1} \mathsf P(X=k)\mathsf P(Y=z-k) & ...


0

You will need two results: The rank of a matrix is inferior or equal to its dimensions. In your case $rank(Y)\le \min (m,t)$. So you will need $t\ge m$. A matrix has rank $m$ if it has a non-zero minor of order $m$ and all minors of strictly superior order are zero. Therefore, your matrix $X$ has a non-zero minor of order $m$. If your sample $Y$ hits the ...


1

It should be $\sum_{n=0}^\infty P(X=2n)=\sum_{n=0}^\infty F(2n)-F(2n-1)=\sum_{n=1}^\infty1-2^{-2n-1}-1+2^{-2n}=\sum_{n=1}^\infty2^{-2n-1}=\frac{1}{2}\sum_{n=1}^\infty4^{-n}=\frac{1}{8}\sum_{n=0}^\infty4^{-n}=\frac{1}{6}$


0

It can be proven fair easily using trace properties that $$ \sum_{i=1}^{n} (z_i - \bar{z})^T \Sigma^{-1} (z_i - \bar{z}) = n \ tr[\Sigma^{-1} \frac{1}{n} \sum_{i=1}^{n} (z_i - \bar{z})^T (z_i - \bar{z})] = n \ tr[\Sigma^{-1} \Sigma^{*}] $$ Putting in $\Sigma = \Sigma^{*}$, yields the desired result.


0

The sum of n independent Bernoulli random variables, all with the same p, is a binomial random variable with parameters n and p. For the case $n=4$, here are a few of the probabilities in the distribution: $P(Y = 0) = (1-p)^4,$ as you already know. $P(Y = 1) = 4p(1-p)^3,$ as in the hint from @Alex. You must take into account all of the sequences 1000, ...


0

Since the system fails as soon as the first component fails, we are looking for the minimum of the lifetimes of the seven components, call it $Y=\min(X_1,X_2\ldots,X_7)$. The cumulative distribution $G$ for $Y$ is $$G(y)=\Pr(Y\le y)$$$$=1-\Pr(Y>y)$$$$=1-\Pr(X_1>y,X_2>y,\ldots,X_7>y)$$ $$=1-[\Pr(X_1>y)\Pr(X_2>y)\ldots\Pr(X_7>y)]$$ ...


0

You want the expected time until the earliest component failure, of seven i.i.d. components.   This is the seventh least order statistic. $$\begin{align} \mathsf E[X_{(7)}] & = \binom{7}{1}\int_1^\infty x\cdot f_{X}(x)\cdot (1-F_X(x))^6 \operatorname d x \\ & = 7\int_1^\infty x \cdot\frac {3}{x^4}\cdot \left(\int_x^\infty \frac ...


1

$Y=0$ if and only if $X_1=X_2=X_3=X_4=0$. Since the four random variables are independent, the probability this happens is $$\Pr[X_1=0]\cdot\Pr[X_2=0]\cdot\Pr[X_3=0]\cdot\Pr[X_4=0]$$ which is equal to $$ (p^0(1-p)^{1-0})\cdot(p^0(1-p)^{1-0})\cdot(p^0(1-p)^{1-0})\cdot(p^0(1-p)^{1-0}) = p^0(1-p)^{4} $$


0

$X$ is binomial with $n = 20$ and $p = .01$ in the first part. Then $a = P(A) = P(X \ge 2)$ is the probability of an event awaited in a geometric distribution. If the 5th hour is the first time $A$ occurs, then you need to find the probability of the following sequence of independent events: $$A_1^c \cap A_2^c \cap A_3^c \cap A_4^c \cap A_5.$$ Can you ...


0

I'm not sure about the variance, but here's an approach for the mean. For $i=1,2,...,n$, let $X_i=1$ if there is a record set on day $i$, and $X_i=0$ if there isn't a record. For $X_i=1$, we must have that the value on day $i$ is the highest of the first $i$ days. So $P(X_i=1)=\frac1i$. So $E[X_i]=\frac1i$. Since expectation is linear we have ...


0

I'm pretty sure you are supposed to use the normal approximation to the normal, as suggested by @Karl. I hope you used continuity corrections to give that the best chance to work. Here are exact binomial answers from R statistical software to compare with your approximate answers: dbinom(100, 100, .9) # (a) 2.65614e-05 # ...


0

Under $H_0,$ the random variable $X$ has the distribution $Binom(4,.2),$ so the PDF (or PMF) which gives $P(X = k),$ for $k = o, 1, 2, 3, 4,$ can be computed as below (unless you know about R statistical software, ignore the first two rows below; they are just a quick way to make the table). k = 0:4; pdf = dbinom(k, 4, .2) cbind(k, pdf) k pdf 0 ...


0

The average probability of a repair $R$ is 1/3 in one year. The probability of n repairs in the year is a Poisson distribution $$P(n) = \frac {R^n e^{-R}}{n!} $$ Assume each machine is independent, so the repair cost of two machines is twice the repair cost of one. The repair cost is uniformly distributed between 0 and 4000, so the expected average cost of ...


0

Since there are 4 cards, there are 4! = 24 possible orderings of the cards. You lose in the following cases: 1) Fire is the first card. There are 3! = 6 ways for this to occur. 2) Wind is the first card and Fire is the second. There are 2! = 2 ways for this to occur. Hence, the probablity of losing = $\frac{6+2}{24}=\frac{1}{3}$ So your answer is ...


2

I'm going to hazard a guess that where you wrote $P(U + (V \min t) \leq s) \implies P(U + V \leq s; V < t)$, you meant "$=$". If the two probabilities are indeed equal, it seems likely that it's because the two events $[U+\min\{V,t\}\le s]$ and $[U+V\le s\ \&\ V<t]$ are equal, or at least differ by an event whose probability is $0$. There are two ...


1

For (a): $P(A \Delta B) = P((A \setminus B) \cup (A \setminus B)) = P(A \setminus B) + P(B \setminus A) = P(A \setminus (A \cap B)) + P(B \cap A^c) = P(A) - P(A \cap B) + P(B \cap A^c) = \frac 1 6 - \frac 1 {12} + \frac 1 {18} = \frac 5 {36}$.


0

Let me Google that for you ... http://en.wikipedia.org/wiki/Merge_sort#Analysis



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