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37

Hint: The probability that an equal number of tails and heads appear is $\large{{2k \choose k} \frac{1}{2^{2k}}}$ The two remaining outcomes (that there are more heads than tails or more tails than heads) are equally likely.


27

I think this is a slightly more intuitive way of looking at the question: Suppose $f (x)=k$. Once we have chosen $k$, there are $200$ possible values for $g (x)$, one of which is $k$, hence we get an answer of $\frac{1}{200}$.


13

Hint: Fair coin $\implies$ Probability of tails occurring more $=$ probability of heads occurring more $= p$, say. Probability of exactly equal number of heads and tails $=1-2p$. Can you find this one?


13

is it possible that the event has in fact zero probability? Yes. For a situation where this always happens, assume that one observes a random number $x$ drawn from the uniform distribution on $(0,1)$. Then the probability to observe $x$ is zero. (Proof: For every interval $I\subseteq(0,1)$ the probability to observe a number in $I$ is the length of ...


12

Let me get you started: if $(X,Y)$ is i.i.d., $(Y,X)$ is distributed like $(X,Y)$ hence $Y-X$ is distributed like $X-Y$ hence $E(u(Y-X))=E(u(X-Y))$ for every $u$ such that the expectations exist. For $u(t)=t^3$ (or any odd function), this yields $E((X-Y)^3)=E((Y-X)^3)=-E((X-Y)^3)$ hence... Thus, the answer is "Yes, provided $E(|X|^3)$ is finite" (otherwise ...


12

Assuming the selections are independent and uniform, there are $100 \times 200$ possible equiprobable pairs of which $100$ can be a match, so the probability is ${100 \over 200 \cdot 100 }$.


12

Start with $x_1=0$ and then just let $x_{n+1}=1$ whenever $f(n)/n\leq p$ and $x_{n+1}=0$ whenever $f(n)/n> p$. It's easy to prove by induction that we will always have $\left|\frac{f(n)}n-p\right|\leq\frac 1n$ and so $f(n)/n\rightarrow p$.


11

The proof of this result has nothing to do with probability theory in general or with Borel-Cantelli lemmas in particular, it is only plain-old-deterministic real analysis: Let $(x_n)$ denote a real valued sequence such that $0\leqslant x_n\lt1$ for every $n$, then $\prod\limits_n(1-x_n)=0$ if and only if $\sum\limits_nx_n$ diverges. Proof: If ...


10

Looking at the final balls is the way to go, as you can regard drawing all $60$ balls as choosing one of the equally probable permutation. Working from the back, you can also ignore a colour once it has been drawn. The probability that the last ball is blue and that the last green comes after the last red is $\dfrac{30}{10+20+30}\times \dfrac{20}{10+20} ...


10

This is false, when $\lim_{n \to \infty} p_n$ doesn't exist. For instance, consider $p_{2n-1} = \dfrac1{2^n}$ and $p_{2n} = 1-\dfrac1{2^n}$. This gives us $\sum_{n=1}^{2N} p_n = \sum_{n=1}^{2N} (1-p_n) = N$, whereas $$\sum_{n=1}^{2N} p_n(1-p_n) < \sum_{n=1}^N \dfrac1{2^{n-1}} < 2$$ However, the claim is true when $\lim_{n \to \infty} p_n$ exists. ...


9

Assuming that by "number" you mean "integer," and by "random" you mean "uniformly at random," AND ASSUMING THAT THE CHOICES ARE INDEPENDENT, then the desired probability is $$\sum_{k=1}^{200} \Pr[X = k]\Pr[Y = k] = \sum_{k=1}^{100} \frac{1}{100} \cdot \frac{1}{200} + \sum_{k=101}^{200} 0 \cdot \frac{1}{200} = \frac{1}{200}.$$


8

The problem is the claim that all $x^2$ values between $0$ and $4$ are equally likely. That's just not true, except in the sense that each one of them has probability zero. But in any more reasonable sense, it fails. In particular, if all were equally likely, then the probability that the square was between 0 and 2 would be the same as the probability that ...


8

Wouldn't this be needlessly complicating things? Stirling formula tells you that $$\Gamma(x+1/2)\sim(2\pi)^{1/2}(x+1/2)^{x}e^{-x-1/2}$$ and that $$ x^{1/2}\,\Gamma(x)\sim(2\pi)^{1/2}x^{x}e^{-x} $$ hence $$ ...


8

Write $A \setminus B$ as $A \cap B^{c}$.


8

Since $\mathbb E[X^2] = \mathbb E[X\mathbb E[Y|X]] = \mathbb E[\mathbb E[XY|X]] = \mathbb E[XY]= \mathbb E[\mathbb E[XY|Y]] = \mathbb E[Y\mathbb E[X|Y]] = \mathbb E[Y^2], $ observe that $$ \mathbb E[(X-Y)^2]=\mathbb E[X^2+Y^2-2XY]=0, $$ which implies that $X$ and $Y$ differ on a set of measure zero. For the weaker condition where $X$ and $Y$ are ...


8

Note that the integral can be written as $$ \int_0^\infty \int_x^\infty f(t)\,\mathrm dt\,\mathrm dx=\int_0^\infty \int_0^\infty f(t)\mathbf{1}_{t\geq x}\,\mathrm dt\,\mathrm dx $$ and using Fubini we get $$ \int_0^\infty \int_0^\infty f(t)\mathbf{1}_{t\geq x}\,\mathrm dx\,\mathrm dt=\int_0^\infty \int_0^t f(t)\,\mathrm dx\,\mathrm dt. $$ Here ...


8

There is a difference between mutually exclusive events and independent events (indeed, a very strict difference in the sense that a couple of non-trivial events cannot be both mutually exclusive and independent). Events $A$ and $B$ are mutually exclusive if it is impossible for both of them to occur simultaneously. To borrow from your example, it is ...


8

I'll try to give you an example that shows why it is very useful. Suppose that the random variable $X$ follows the binomial distribution with parameters $n$ and $p$. Then the expected value of $X$ is given by $$ \operatorname EX=\sum_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k}. $$ But we know that $X=\sum_{j=1}^nY_j$, where $Y_1,\ldots,Y_n$ are independent and ...


7

With such a large number of trials and with $p = 0.5$, a normal approximation to the binomial distribution would also work. if $X \sim \mathrm{Binomial}(n = 10^6, p = 0.5)$, then $$\Pr[X = n/2] = \Pr\left[\frac{X - np}{\sqrt{np(1-p)}} = \frac{n/2 - np}{\sqrt{np(1-p)}}\right] \approx \Pr[-1/\sqrt{n} \le Z \le 1/\sqrt{n}]$$ using continuity correction, where ...


7

Then can you help me to show that $\alpha$ is in fact stopping time? Well, not very easily I am afraid, since $\alpha$ is not always a stopping time for the filtration $(\mathcal F_t)_{t\geqslant0}$ defined by $\mathcal F_t=\sigma(X_s;0\leqslant s\leqslant t)$. For a counterexample, consider some Bernoulli random variable $U$ such that $P(U=1)=\frac12$ ...


7

Your final expression: $$a_k = \frac{1}{5} \cdot\left( \frac{25}{36} \right)^k $$ This correctly gives the probability that your first 6 will occur on the $2k$th roll. So, $a_1$ is the probability that your first 6 is on roll 2; $a_2$ is the probability that your first 6 is on roll 4; etc. In order to find the probability $p$ that your first 6 is on any ...


7

The measure $P_X$ (not $P_x$) is the image measure of $P$ by $X$, defined by the identity $P_X(B)=P(X^{-1}(B))$ for every $B$ in $\mathcal B(\mathbb R)$ (and $X$ being a random variable is exactly the hypothesis one needs to be sure that $P(X^{-1}(B))$ exists). The fact that, for every measurable function $u$ suitably integrable, $$ \int_\Omega u(X)\,\mathrm ...


7

The probability is $0$, assuming that each point is chosen with uniform distribution on the circle. For a right-angled triangle you need two of the points to be the two endpoints of a diameter. The probability for that is obviously $0$, since there isn't any range of possibilities, and the distribution is continuous. For anyone who is not satisfied so ...


7

Note that $$\bigcup_{n=1}^\infty A_n \supset \bigcup_{n=1}^\infty A_n \cap \bigcup_{n=2}^\infty A_n \cap \ldots \cap \bigcup_{n=k}^\infty A_n \cap \ldots = \bigcap_{k=1}^\infty \bigcup_{n=k}^\infty A_n = \limsup_{n\to\infty} A_n$$ And use $A\subset B \Rightarrow P(A) \le P(B)$


7

Let $S_{100} = \sum_{k=1}^{100}X_k$ denote the winnings after 100 bets. Then $$E(S_{100})= 100E(X)= 25$$ and $$var(S_{100})= \sum_{k=1}^{100}var(X)= 100(27/16)\approx 168.75.$$ As a sum of $100$ independent random variables, $S_{100}$ is approximately normally distributed with mean $25$ and standard deviation $\sqrt{168.75} \approx 12.99$. The ...


7

Suppose that I have a set $S$ of white balls. The calculation $$\binom{n}k\binom{n-k}\ell$$ counts the ways to perform the following operation: pick $k$ of the balls in $S$ and paint them red, then pick $\ell$ of the remaining white balls and color them blue. I could get the same results by first picking any $k+\ell$ balls, then choosing $k$ of those to ...


7

For concreteness, assume that we have a biased coin that has unknown probability of landing "heads" and we want to give a $95\%$ confidence interval for $p$. Rrepeat the experiment of tossing the coin $n$ times. Let $\bar{X}$ be the sample proportion of heads. We will use $\bar{X}$ to estimate $p$. Note that $\bar{X}$ has mean $p$ and variance ...


6

$$T=ZX+(1-Z)Y\implies E(\mathrm e^{\mathrm itT})=pE(\mathrm e^{\mathrm itX})+(1-p)E(\mathrm e^{\mathrm itY})$$


6

It turns out the numerical findings about $\mathbb{E}[N_3]$ by David E is not a coincidence, it is exact! $$\mathbb{E}[N_3] = \frac{1}{1-\sin(1)}$$ Let $X_1, X_2, \ldots $ be a sequence of random variables. For each $n$, we will assume $X_n$ take value from the set $\langle n \rangle \stackrel{def}{=}\{ 1, 2, \ldots, n \}$ with uniform probability. After ...



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