# Tag Info

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 )

2

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 ...

2

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 ...

1

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) + ... 1 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}) &= ...

1

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. $$1 Let X be the number of people with disease. It's clear that X has binomial distribution with n=10000 and p=\frac{1}{1200}. We can calculate the exact probability using the formula$$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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