I am asking to use central limit theorem to solve this quesion. In an election between two candidates, A and B, one million individuals cast their vote.  Among these, 2000 know candidate A from her election campaign and vote unanimously for her. The remaining 998000 voters are undecided and make their decision independently of each other by flipping a fair coin.  Approximate the probability pA that candidate A wins up to 3 significant figures. I got my mean = np = (1000000-2000)  * 0.5 = 499000, my sd =$\sqrt {np*(1-p)}$= $\sqrt {998000*0.5*(1-0.5)}$, then apply CLT, my new sd = $\frac{sd}{\sqrt{998000}}$, and X = $\frac{1000000}{2}$+1-2000 = 498001. Then apply normal distribution. Am I right?
121973
I am asked to solve the following question using central limit theorem. In an election between two candidates, A and B, one million individuals cast theirvote. Among these, 2000 know candidate A from her election campaign and vote unanimously for her.The remaining 998000 voters are undecided and make their decision independently of each other byflipping a fair coin.Approximate the probability pA that candidate A wins up to 3 significant figures.  It's easy to solve directly. PA = $\frac{0.5(10000-2000)+2000}{10000}$ = 0.501. However, I am quite confused about how to solve this problem by central limit theorem.
102392
I'm learning applications on Central Limit Theorem and got really confused with a few points. According to this , the procedure to apply CLT usually goes like this:   So if SD is the population standard deviation, how are we gonna get it?? Isn't the whole population standard deviation what we eventually calculate by applying CLT and analyzing a sample of the whole population? How come the population standard deviation become a prerequisite??   Please tell me this tutorial is wrong.   I think the SD actually refers to the standard deviation of a sample (of some size n), which we can actually get easily. For example:    sample the whole population with a sample size of n (e.g. randomly select 10000 users from the whole population of 10 billions) calculate the mean of the 10000 measurements. calculate SE = SD / sqrt(10000), where SD is the standard deviation of the 10000 measurements, instead of the standard deviation of the whole population of 10 billions.    This explanation would make much more sense.   Any thoughts?
123816
A person randomly chooses a coin with probability 0.5: Tossing coin 1 yields a head with a probability P(X1=H)=.3 (and tail with P(X1=T)=.7). Tossing coin 2 yields a head with a probability P(X2=H)=.6 (and tail with P(X2=T)=.4). You earn 1 dollar if you correctly guessed the coin and 0 dollars otherwise. Design the optimum decision rule and estimate your average earning.  Can anybody explain me ,how should i go about solving this problem?
50305
Let's say I have one point that will be taken randomly from a normal distribution with mean $\mu_1$ and standard deviation $\sigma_1$.  Let's say I have another point that is taken much in the same way from another normal distribution with mean $\mu_2$ and standard deviation $\sigma_2$;.  How can I compute the probability, given $\mu_1$, $\mu_2$, $\sigma_1$, and $\sigma_2$, that my first point will be larger than the second?  I am sort of interested in the reasoning behind an "analytic" answer (or as analytic as you can possibly get with the normal distribution, which isn't that much), but I am more importantly looking for an algorithm of computing this probability, as it will be used in a simulation/model.  Does anyone know where I could get started on reasoning through this?  Note: For actual computation, having a table of values of the % of the curve within a given multiple of the standard deviation is feasible in my situation.
123783
There is something that I am doing wrong in the exercise below and I would appreciate some help figuring it out.  Let $X_1, X_2, \ldots, X_5$ be a random sample from a $\Gamma \left(3,3 \right)$ distribution . Define $W=\sum_{i=1}^5 X_i $.    Obtain the distribution of $W$ and of $2W/3$ Find $c_1$ and $c_2$ such that $P \left( c_1 &lt; W &lt;c_2 \right)=0.95 $     Since $W$ is the sum of iid gamma variables, $W \sim \Gamma \left(15,3 \right)$ and now a straightforward transformation argument shows that $Y=\frac{2}{3} W \sim \chi^2 \left(30 \right) $. This completes question 1.  Now for question 2, my thoughts were to transform these probabilities using the result from question 1, namely that $\frac{2}{3} W \sim \chi^2 \left(30 \right) $. Hence, unless I have made a horrible mistake here:  $$P \left( \frac{2}{3} c_1 &lt; Y &lt; \frac{2}{3} c_2 \right)=0.95 $$  We can find the constants from a chisquare table for $30$ degrees of freedom. In my case, they are $16.791$ for the lower bound and $46.979$ for the upper one.   Then if we equate $\frac{2}{3}c_1=16.791$ and solve for $c_1$ we should be able to find the constant for the above gamma confidence interval, right?   The problem is that according to $R$, $P \left( W&lt;c_1 \right) =1 $, instead of  the desired $0.025$ which means I have gone wrong somewhere. I do not think it is in question though $1$ as I have verified the result many times.   Thank you.
51859
Problem: This is problem 33 of section 2.6 (conditioning) in Bertsekas, Tsitsiklis intro to probability text.  We are given that a coin has probability of heads equal to p and tails equal to q and it is tossed successively and independently until a head comes twice in a row or a tail comes twice in a row. What isthe expected value of the number of tosses?  Solution:  Let $X$ be a random value that tracks the number of tosses. In addition, let $H_k, T_k$ be the event that a head or tail comes up on the kth flip respectively. The solution I am looking at involves using total expectation theorem.  I outline the following steps in the solution.  We can partition the sample space into $H_1$ and $T_1$, allowing us to invoke the total expectation theorem. Hence,  $E[X] = pE[X|H_1] + qE[X|T_1]$  Again by total expectation theorem, we can utilize another partition to obtain  $E[X|H_1] = pE[X|H_1,H_2] + qE[X|H_1,T_2] = 2p + q(1 + E[X|T_1])$  $E[X|T_1] = qE[X|T_1,T_2] + pE[X|T_1,H_2] = 2q + p(1+E[X|H_1])$  Stuck: The solution follows up by saying that we can combine the above two relations and use the fact that p+q = 1 to obtain  $E[X|T_1] = \frac{2+p^2}{1-pq} , E[X|H_1] = \frac{2+q^2}{1-pq}$  I have been staring at this for a while now, but I cannot seem to see the steps of the computation. I'd be very grateful if someone could fill in the details regarding the computation. Thank you.
16735
For each positive even integer $n$, set   $$P_n = \displaystyle {{n} \choose {\frac{n}{2}}}\frac{1}{2^n}.$$  Show that $\displaystyle \lim_{n \to \infty} P_n$ exists and determine its value.   Here's what I have so far: each $P_n$ can be thought of as the probability of tossing a fair coin $n$ times, and obtaining exactly $\frac{n}{2}$ heads. My expectation is that $P_n \to 1$, since in any large trial, one would expect to record just about as many heads as tails. But I'm not sure how to mathematically justify this hunch.  Hints or solutions are greatly appreciated.
39215
What is wrong with the following sentence:      The Central Limit Theorem  implies that, as the sample size grows, the error distribution approaches normality.   Am I correct by saying that it should in stead state the MEAN of the sample error approaches zero as sample size grows?
22292
Yesterday night, while I was trying to sleep, I found myself stuck with a simple statistics problem. Let's imagine we have a "magical coin", which is completely identical to a normal coin but for a thing: every time you toss the coin, the probability to get a head halves. So at t = 1 we have 50:50, then 25:75, then 12.5:87.5 and so on. At t = ∞ we are going to have 0:100.  My question is: if I toss this magical coin infinite times, can I say I am sure I am going to get at least one head? On one hand, I thought, the law of large numbers states that if an event is repeated infinite times, every state that is possible is bound to happen. On the other side, however, at t = ∞ the probability to get a head is zero.  Surely the solution of the problem is fairly easy, so what am I doing wrong?
73966
Here is the problem setup:  Two people play a game of throwing balls at a pin. Sally hits the pin 70% of the times she throws, and Bill hits the pin 40% of the times he throws. The game is won by the person who hits the target first. If Bill throws first each time, find the probability he will win.  With simulation, I get Bill's probability of winning to be around 0.52 (rounding to two decimal places). How to solve this problem analytically?
96192
I am modeling a random variable ($Y$) which is the sum of some ~15-40k independent Bernoulli random variables ($X_i$), each with a different success probability ($p_i$). Formally, $Y=\sum X_i$ where $\Pr(X_i=1)=p_i$ and $\Pr(X_i=0)=1-p_i$.  I am interested in quickly answering queries such as $\Pr(Y&lt;=k)$ (where $k$ is given).  Currently, I use random simulations to answer such queries. I randomly draw each $X_i$ according to its $p_i$, then sum all $X_i$ values to get $Y&#39;$. I repeat this process a few thousand times and return the fraction of times $\Pr(Y&#39;\leq k)$.  Obviously, this is not totally accurate (although accuracy greatly increases as the number of simulations increases). Also, it seems I have enough data about the distribution to avoid the use simulations. Can you think of a reasonable way to get the exact probability $\Pr(Y\leq k)$?  p.s.  I use Perl &amp; R.  EDIT  Following the responses I thought some clarifications might be needed. I will shortly describe the setting of my problem. Given is a circular genome with  circumference c and a set of n ranges mapped to it. For example, c=3*10^9 and ranges={[100,200],[50,1000],[3*10^9-1,1000],...}. Note all ranges are closed (both ends are inclusive). Also note that we only deal with integers (whole units).  I am looking for regions on the circle that are undercovered by the given n mapped ranges. So to test whether a given a range of length x on the circle is undercovered, I test the hypothesis that the n ranges are mapped randomly. The probability a mapped range of length q&gt;x will fully cover the given range of length x is (q-x)/c. This probability becomes quite small when c is large and/or q is small. What I'm interested is the number of ranges (out of n) which cover x. This is how Y is formed.   I test my null hypothesis vs. one sided alternative (undercoverage). Also note I am testing multiple hypothesis (different x lengths), and sure to correct for this.
101574
I'm stuck on the following probability problem: parents keep having children until they have one girl, at which point they stop; and babies are girls with probability 0.49. If we select a child uniformly at random (from the entire population of children) what's the probability he or she has exactly one sibling?  If a couple has only $2$ children it means that they had first a boy and then a girl. So why the probability asked is not $0.51 \cdot 0.49$? What I am missing?   It seems that the correct answer is $2 \cdot 0.51 \cdot (0.49)^2$, but I don't know how to get this result. Thanks for your help.
72323
Ok, I know this is wrong, but I want someone to tell me why.   Let's take a normal heads tails example of a fair coin. The probability of getting head = 1/2. And I write this is because, either it will be heads, or not. Hence two cases that makes it 1/2.  Now, I know I can't do this to all the cases. For example, the probability of getting a 2 when I roll a dice. I know the answer is 1/6 but why can't I do, either the outcome will be 2 or not. And in that case, my probability is 1/2.
21223
Here is the setup: Bob runs an experiment: he flips a coin N times (between 0 and +$\infty$). The coin has a probability p of landing on heads. Bob starts with zero points. For each head, Bob scores a point. Now, Bob goes to Alice and tells her p and that the score is zero, and ask her the probability that he flipped the coin N=n times. The case that make it special for me is the addition of the possibility of zero draws, and I'm stuck with it. Here is my attempt at solving this: N is the random variable counting the number of flips, S is the random variable counting the score. We want to know $P(N=n|S=0)$. $$ P(N=n|S=0) = \frac{P(S=0|N=n) \cdot P(N=n)}{P(S=0)} $$ Now $P(S=0|N=n)= (1-p)^{n}$ so that's good. Also, $P(S=0) = P(N=0) + \sum\limits_{n=1}^\infty (1-p)^{n}P(N=n) $ But now I'm stuck since I don't know what do with that $P(N=n)$. Now you could argue that since we know nothing about the distribution of N, the most reasonable assumption is that they are all equally likely. Then the substitution occurs as: $$ P(N=n|S=0) = \frac{P(S=0|N=n) \cdot P(N)}{\frac{1}{p}\cdot P(N)} = p(1-p)^{n} $$ This feels like cheating. Not because of the limit, this looks legit cf replace $\infty$ with $N_{max}$ and take the limit at the end, but because of the assumption on $P(N)$ being uniform. While keeping $N_{max}$ this is good, but it seems shady when taking the limit. Also, what do I know if it is uniform ? Am I doing something wrong here ?
76514
folks,  i am new to this forum and not a math expert. so please bear with me if am asking silly questions.   The question is "probability of getting 50 heads from tossing a coin 100 times".  So the answer for this is, I guess, ${100 \choose 50} (2 ^{-100})$.  So all am trying to get is easier way to calculate ${100 \choose 50}$, or another approach to the parent problem only.  Thanks all, appreciate that.  raj
22029
$$P(X_1 &lt; X_2 &lt; X_3)= \frac{(1-p_1)(1-p_2)p_2p_3^2}{(1-p_2p_3)(1-p_1p_2p_3)}$$   To be frank I do not know where to start with this question, I would like an idea to get me going, or better yet an answer. This was on an earlier exam, I would like to be ready if something like this were to come up again tomorrow. What makes this a little bit more difficult than usual is the fact that geometric distribution is  a discrete variable..
13049
Let $X_1,X_2,\ldots,X_n$ be a random sample of Bernoulli distribution with parameter $\frac{\theta_1}{\theta_1+\theta_2}$ and $Y_1,Y_2,\ldots,Y_n$ be a random sample of geometric distribution with parameter $\theta_1+\theta_2$. If the two samples are independent of each other, how can I calculate the limiting distribution of $\dfrac{\sum_{i=1}^n X_i}{\sum_{i=1}^n Y_i}$ ?