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3

No, the statement is not true. Let for example $Ω=\{1,2,3,4\}$ with $p(ω)=1/4$ for each $ω$ and let $X=\{1,2\},\, Y=\{2,3\}, \, Z=\{3,4\}$. Then $$P(X\cap Y\cap Z)=P(\emptyset)=0$$ but you can check that $X,Y,Z$ are pairwise independent with $P(X)=P(Y)=P(Z)=\frac12>0$.


2

$(n-j)!\over {n!}$. That is because $n-j$ people can sit in $(n-j)!$ ways. ( $j$ people already sat in $j$ seats. So, rest $(n-j)$ people sit in $(n-j)!$ ways )


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I will use $t$ as the variable instead of $\lambda$. The required expectation is $$\left(E(\exp(\frac{tX_1}{n})\right)^n,$$ and $$E(\exp(\frac{tX_1}{n}))=(1-p)+pe^{\frac{t}{n}}.$$


2

The answer you got - 0.246 is the probability of getting 'exactly' 5 heads. Your intuition gives the 'Expectation' E(x). When 10 coins are tossed, the Expectation is that you get 5 heads. Your intuition makes an average of all the cases while the solution takes only the cases where number of heads is 'exactly' 5. Let me give you another problem- What is the ...


2

You are describing the multi-armed bandit problem. You have $N$ slot machines (decks), each with some unknown expected payoff (win %). You want to maximize your payoff, which demands a careful balance between exploration (gathering data about each slot machine/deck) and exploitation (using the slot machine/deck that appears to be best so far). There are a ...


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Edit: The original question asked about independence, and was answered by the example below. This example also settles the modified question about covariance, since $\text{Cov}(XY,Z)\ne 0$. Toss a fair coin twice. Let $X=1$ if we have head on the first toss, and $0$ otherwise. Let $Y=1$ if we have head on the second, and $0$ otherwise. Let $Z=1$ if the ...


2

\begin{align}P(WW)&=\sum_{i=1}^{11}P(WW\mid bin=i)P(bin=i)=\sum_{i=1}^{9}\frac{\dbinom{10-(i-1)}{2}}{\dbinom{10}{2}}\frac{1}{11}\\[0.2cm]&=\frac{1}{11\cdot45}\sum_{i=1}^9\dbinom{11-i}{2}=\frac{1}{11\cdot 45}\sum_{i=0}^{8}\dbinom{2+i}{2}\\[0.4cm]P(BB)&=\sum_{i=1}^{11}P(BB\mid ...


1

The probability is $$\ _{(n-k)}C_r\over \ _nC_r$$ choosing $r$ balls from $(n-k)$ [favourable cases] divided by choosing $r$ balls from $n$ [sample space].


1

Let $E$ be the event the workout ends early. Then $$\Pr(E\mid W^c)=\frac{\Pr(E\cap W^c)}{\Pr(W^c)}.$$ You have calculated $\Pr(W^c)$. We now need to find $\Pr(E\cap W^c)$. The event $E\cap W^c$ can happen in two ways: (i) we drink coffee and the workout ends early, or (ii) we drink the commercial drink and the workout ends early. The probability of (i) is ...


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In short, you erroneously divided by $\Pr\{\text{sum is 9}\}$ when calculating $\Pr\{\text{sum is 9 and exactly one die lands on 6}\}$. $$ \begin{align*} \Pr\{\text{sum is 9 and exactly one die lands on 6}\} & = \Pr\{(6,3),(3,6)\} \\ & = \Pr\{(6,3)\} + \Pr\{(3,6)\} \\ & = \left(\frac{1}{2}\right)\left(\frac{1}{6}\right) + ...


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The first mistake is when you calculated the probability that exactly one die lands on $6$, as you included the case when both dice are $6$ twice (first when the loaded die is $6$, and secondly when the fair die is $6$) - so you need to subtract the case when they were both $6$ two times:- $$\begin{align}P(\text{exactly one die lands on 6}) &= ...


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Let's imagine a case with $4$ rolls where we desire $2$ heads (MathJax diagrams only go so far...) $$\newcommand{\mychoose}[2]{\bigl({{#1}\atop#2}\bigr)}$$ $$\begin{array}{ccccccccccc} & & & & & & & H & & & & & \\ & & & & & & \swarrow & & \searrow & \\ & ...


1

Yes, that's enough. Remember that $\limsup_n a_n = \lim_n\{\sup_{k\ge n} a_k\}$ for any sequence $\{a_n\}$ of real numbers. If the limsup is bounded above by some $A$, then there exists $N$ such that $$ \sup_{k\ge N} a_k \le A +1. $$ But this implies that $a_k\le A +1$ for all $k\ge N$ ( since the sup is an upper bound on every term ), so $A+1$ is a uniform ...


1

If the chance that $A$ throws the first $6$ is $p$, that can happen if he throws a $6$ on his first turn or if nobody throws a $6$ on their first turn and $A$ is first to throw a $6$ after that, which is $p$. So $p=\frac 16 + (\frac 56)^3p$ Now the chance that $B$ is first after $A$ is exactly $p$ and the chance that $C$ is first after that $B$ throw is ...


1

In the first question, A might throw several $6$ before B gets any. In the second question, that is not allowed. To solve the first problem, let $a$ be the probability A is first to throw a $6$. This can happen in two ways: (i) A throws an immediate $6$ or (ii) A misses, then B misses, then C misses, but A ultimately is first. That gives us the equation ...


1

To expand on the method I sketched in the comments: There are $6^3=216$ possible ordered triples. Consider first those triples with no duplicates. There are $\binom 63 = 20$ unordered such triples. As there are $3!=6$ permutations on three letters, there are $120$ ordered triples with no duplicates. Note that there is exactly one way to order such a ...


1

This random variable has the binomial distribution. Hence, the probability that an island with $10000$ inhabitants has precisely $8$ people born with that particular disease is given by $$ {10000\choose 8}\biggl(\frac1{1200}\biggr)^8\biggl(1-\frac1{1200}\biggr)^{(10000-8)}\approx0.1387. $$


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Let $X$ be the number of people with disease. It's clear that $X$ has binomial distribution with $n=10000$ and $p=\frac{1}{1200}$. We can calculate the exact probability using the formula $$P(X=8) = {10000\choose 8}\biggl(\frac{1}{1200}\biggr)^8\biggl(1-\frac{1}{1200}\biggr)^{(10000-8)}$$ The exact calculation of the above quantity is hard. We can ...


1

Let's look at Problem 3.5 again: 3.5 An urn contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement, what is the probability that the first 2 selected are white and the last 2 black? Here, the only thing given to you is the situation that there are $6$ white and $9$ black balls and that $4$ balls are going to be ...


1

The other responder has given an elegant solution, but here is another way by enumerating it a little bit. Probability for any one case could be calculated such as this P(Picking a Head from the Table)*P(2 books being Head)*P(1 book being a head) for the first case. The three books could be H H , H $$= 1.\dfrac{{5\choose2}}{{11\choose2}}.\frac{7}{11} ...


1

The equation as stated without additional assumptions on the independence of $A$ and $B$ is false. Consider the following counterexample. Consider the experiment where we flip two fair coins in sequence. We have the equiprobable sample space $\{HH, HT, TH, TT\}$ Let $A$ be the event that the first coin is a head. $Pr(A)=0.5$ Let $B$ be the event that ...



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