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7

I think that's a misunderstanding. The passage you quote is a bit informally phrased. By "the mean time to failure of some disk", they don't mean the mean time to failure of each individual disk but the mean time it takes until any one of the $100$ disks fails. If the failures form a Poisson process, then having $100$ disks instead of $1$ disk will increase ...


4

Compare it to lottery. If you own one ticket, the probability to win is very low. If you own 100 tickets it's much more likely that you win (something). Likewise, the probability that one disk fails is quite low but the probability that any of the 100 disks fails is higher, and therefore the mean time to failure is lower.


4

If $p_n$ is the probability that the $n$th roll is a six, the generating function for this sequence is $$\phi(s)=\sum_{n\geq1} p_n s^n={120s\over(6-s)(5-s)(4-s)(3-s)(2-s)(1-s)}.$$ A partial fraction expansion gives us the explicit formula $$p_n=\sum_{j=0}^5 {5\choose j}{(-1)^j\over (j+1)^n}.$$ The first few values are $$p_1={1\over 6},\quad p_2={49\over ...


3

The complementary event of $A = $"at least one student speaks Spanish" is that $\bar A =$None (of the three) speaks Spanish. To calculate the chance that no student speaks Spanish, it has to be the case that the three students do not speak Spanish which is $(1-.3)\times (1-.3)\times (1-.3)$ . Hence, $$P(A) = 1-P(\bar A) = 1-(1-.3)^3 = 0.657$$ The reason ...


3

The mean time to failure of some disk is not the mean time to failure of those disks: it is the mean time before at least one of those 100 disks fails. In other words, the time to failure of some disk $T$ is the minimum of the time to failure of all disks, i.e. $$T = \min_{i \leq i \leq N} \{X_i\}.$$ A standard of model is that aging has no memory ...


2

It is better to think about the complement. The opposite of selecting at least 1 Spanish speaker is to pick all three students who do NOT speak Spanish. Can you compute this probability? Then your desired answer is just 100% minus this answer.


1

For $y<0$, obviously $f_Y(y)=0$. For $+\infty>y>4$, $$f_Y(y)=f_X(\sqrt{y})\bigg|\frac{dx}{dy}\bigg|$$ $$=\lambda e^{-\lambda(\sqrt{y}+2)}\frac{1}{2\sqrt{y}}$$ $$=\frac{\lambda e^{-2\lambda}}{2}\frac{e^{-\lambda\sqrt{y}}}{\sqrt{y}}$$ For $0 \le y \le 4$, ...


1

where I don't only demand $r$ red ones, but also exactly $g$ yellow ones and $b$ blue ones. Can I somehow iterate or modify the $HyG$ formula to match this new scenario? Yes.   You want the probability of drawing $r$ from $R$ red, $g$ from $G$ green yellow, and $b$ from $B$ blue, out of all the ways to draw any $r+b+g$ from all $R+G+B$ ...


1

It is not that complicated. The conditional expectation, of random variable given an event, is defined in terms of conditional probability. $$\begin{align}\mathsf E(Z\mid Z\geq 0) ~=~& \int_\Bbb R z~f_Z(z\mid Z\geq 0)\operatorname d z\\[1ex]~=~& \int_0^1 z f_Z(z)\operatorname d z~\Big/~\int_0^1 f_Z(z)\operatorname d z \end{align}$$ Also you have ...


1

Yes, your reasoning is correct. However, I would say that applying the monotone convergence theorem is kind of overkill. What you are really using is that the probability measure $\mathbb{P}$ is continuous from above (i.e. that $A_k \downarrow A$ implies $\mathbb{P}(A_k) \to \mathbb{P}(A)$.) Simply note that $$A_k := \{X>a_k\} \downarrow \{X>0\}$$ ...


1

The expected value is calculated by using $$E[X] = x_1p_1+x_2p_2+\cdots+x_kp_k$$ Case 1 = head occurs The expected value on the die is calculated as follows $$\begin{align}E[X\mid \textsf{Heads }] = ~& 1\cdot \frac{1}{6} + 2\cdot \frac{1}{6} + 3\cdot \frac{1}{6} + 4\cdot \frac{1}{6} + 5\cdot \frac{1}{6}+ 6\cdot \frac{1}{6}\\[1ex] = ~& ...


1

To solve this, I would first calculate the expected value of a single die: ${1+2+3+4+5+6\over6}={7\over2}$ So that the expected value of two dice is: ${7\over2}+{7\over2}=7$ Then, the overall expected value is: ${7+{7\over2}\over2}={21\over4}=5.25$ This method is only possible because all dice rolls occur with equal probability (1/6) and both heads and ...


1

Let $T$ be the count of tails on the coin toss (ie $0$ for heads, $1$ for tails).   Thus can partition the space over these two disjoint events, using the Law of Total Expectation.   Because it is a fair coin: $$\mathsf P(T=0)~=~\mathsf P(T=1)\\~~~=~\tfrac 12$$ Let $X$ be the sum of the dice results, and $X_1,X_2$ be the result of the individual ...


1

I am guessing that you are trying to compute variance of $X$. The problem is we did not define $X$. In your lecture note, $X$ seems defined to be waiting time. Waiting time comes in multiple of 3 in this setting. We have $Var[X]=E[X^2]-E[X]^2$ On the other hand, it seems that you are trying to wokr with the number of copies.To avoid confusion, let's ...


1

$A_2=\left\{2,4,6,8,\ldots,20 \right\}$, the even numbers. $$P(A_2)=\frac{10}{20}=\frac{1}{2}$$


1

$S = \{1,2,3,\ldots,20\}$ $A_n = \{x \in S: x \: \text{is a multiple of} \:n\}$ $\:\:\:n=1,2,3,\ldots$ $A_1 = \{1,2,3,\ldots,20\}$ $A_2 = \{2,4,6,\ldots,20\}$ because the products of $n=2$ with the elements in $S$ are $2,4,6,\ldots,20$ up to $20$ because we have the restriction that $x \in S$. So $1 \leq x \leq 20$ $A_3 = \{3,6,9,\ldots,18\}$ and ...


1

As Ian said, $T_n$ for $n>1$ is actually Gamma Distributed with $k=1$. In fact, the Exponential Distribution is a special case with $k=1$. You did mention you were using $T_n$ for the interarrival times, but that isn't the standard notation - $T_n - T_{n-1}$ is the interarrival time, where $T_n$ is simply the total arrival time for $x_n$. I think you're ...


1

Use Borel-Cantelli. Let $n=k\cdot 2m$ and $$A_k:=\left\{X_{2m\cdot k }=1\cap X_{2m\cdot k+1 }=1\cap\ldots\cap X_{2m\cdot k+m-1}=1\right\}$$ Then $A_k$ are independent, and $$ \mathbb{P}(A_k)=\prod_{j=0}^{m-1}\mathbb{P}(X_{2m\cdot k+j }=1)=p^m $$ So: $$ \sum_{k=1}^\infty\mathbb{P}(A_k)=\infty $$ Thus by Borel-Cantelli, infinitely many $A_k$ will happen a.s.



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