A certain police officer stops cars for speeding. The number of red sports cars she stops in one hour is a Poisson process with rate 4, while the number of other cars she stops is a Poisson process with rate 1. Assume that these two processes are independent of each other. Find the probability that this police officer stops at least 2 ordinary cars before she stops 3 red sports cars.  How do you solve it by binomial or by poisson? My initial response was to consider p = 1/5  How do you solve poisson process problems in which you need to calculate the probability of one event before another? In this case, the probability of 2 ordinary cars before 3 sports cars. By poisson, I tried calculate the event of 2 ordinary cars and event of 3 red sports cars. What I can't understand is one event before another logic. Can we do these problems by binomial?
122815
Next weekend you will be participating in 12km cross country race on a mountain.The average time between  two successive wild animal sightings on the mountain is reported to be 5 minutes  (a) What is the probability that you see at least one wild animal in the 11th minute of the     race given that you will see 3 wild animals since the start of the race?  (b) What is the probability that it will take more than a quarter of a hour before you see a wild animal after ten minutes of running?  Now I have attempted (a) , but I don't know whether my thinking is correct. (b) on the other hand makes little sense to me.  My attempt:  (a)$ X~ Poisson(\frac{1}{5})$ and $Y~Exponential(\frac{1}{5})$  $Pr(X&gt;4) = Pr(X&gt;1)$  (I am thinking that this is some variation of the memoryless property)  $Pr(X&gt;1) =Pr(Y&lt;1)$  $Pr(Y&lt;1)= 1 - (e^{(\frac{-1}{5})})$  Perhaps, I was a bit to vague in my attempt of (a). So here goes my attempt of(b).From my understanding of the question , it is asking what it the probability that the time between events(in this case animal sightings) is more than 25 minutes given you have been running for ten minutes  Now from what I know the fact that you have been running ten minutes is irrelevant this is due to memory-less property of the exponential  distribution   so without further ado  I present my attempt at (b)   let $X $ be exponentially distributed random variable with $\lambda = 1/5$  then  $Pr(X &gt;15) = 1 - Pr(X &lt;= 15)$
102300
There are 5 urns, the nth urn contains n − 1 red balls and 5 − n green balls (in total 4 ineach urn). We pick an urn at random and take 2 balls out of it without replacement.  (a) What is the probability the second ball is green?  (b) What is the probability the second ball is green given the ﬁrst is green?  (c) What is the probability we picked either the ﬁrst urn or second urn given both balls are green?  for (a), I did:   P(2nd ball is green) = 5*[{P(i th urn picked) * P(first ball drawn red/ ith urn picked) P(2nd ball drwan green/first ball red)}+ {P(i th urn picked) * P(first ball drawn green/ ith urn picked) P(2nd ball drwan green/first ball green)}]  = 5*[[{1/5 * (n-1)/4 * (5-n)/3 } + {1/5 * (5-n)/4 * (4-n)/3}] = 5* [{(n-1)(5-n)/60} + {(5-n)(4-n)/ 60}]  =[(n-1)(5-n) + (5-n)(4-n)]/12 = (5-n)(n-1+4-n)/12 = (5-n)*3/12 = (5-n)/4  And for b): 5*{P(i th urn picked) * P(first ball drawn green/ ith urn picked) P(2nd ball drwan green/first ball green)} = 5*{1/5 * (5-n)/4 * (4-n)/3} =(4-n)(5-n)/12  c), P(1st urn or 2nd urn/ both are green) = [ [P(1st or 2nd]* (4-n)(5-n)/6 ] / [{P(1st or 2nd]* (4-n)(5-n)/6} + {{P(3rd or 4th]* (4-n)(5-n)/6}] = {(4-n)(5-n)/15} /  2*{(4-n)(5-n)/15} = 1/2  But I am really uncertain about my solutions. If anyone has different solution or easier way to get the answer please let me know! Thanks a lot!
29868
A box contains $731$ black balls and $2000$ white balls. The following process is to be repeated as long as possible.  (1) arbitrarily select two balls from the box. If they are of the same color, throw them out and put a black ball into the box. (We have sufficient black balls for this).  (2) if they are of different colors, place the white ball back into the box and throw the black ball away.  What will happen at last? Will the process stop with a single black ball in the box or a single white ball in the box or with an empty box?  I am unable to decide how to start and in which direction? should we apply probability or what?
13089
I know that $$P\left(\bigcup_{i=1}^{n} A_i \right)$$   is the sum of of the probabilities of all the sample points that are contained in at least one of the $A_{i}$'s. This is the probability of sample points belonging to exactly 1 event, exactly 2 events, ...,exactly $n$ events. WLOG this can be written as $$P\left(\bigcup_{i=1}^{n} A_i \right) = P(A_1) + P(A_1 \cap A_2) + \cdots + P(A_1 \cap A_2 \cap \cdots \cap A_n)$$   But the $A_i$'s are arbitrary and we have to account for that. So there are $n$ possibilities for the first probability, $\binom{n}{2}$ possibilities for the second probability, ..., and $1$ possibility for the final probability. So we add and subtract these to prevent overcounting?
55900
Working my way through the following problem: Problem  Suppose that $E$ and $F$ are mutually exclusive events of an experiment.  Show that if independent trials of this experiment are performed,  then $E$ will occur before $F$ with probability $\frac{ P( E)}{P( E) + P( F)}.$  I have the following come up with the following solution: Solution  Since $P( E^c) = P( F)$ Therefore $\frac{ P( E)}{ P( E) + P( F)} = \frac{ P( E)}{ 1 - P( F) + P( F)} = \frac{ P( E)}{ 1} = P( E)$  But I am unsure if I am able to assume $P( E^c) = P( F)$ as a given? As well, I am particularly confused by the answer in the solution manual which makes it's argument as follows: Solution Manual  If $E$ and $F$ are mutually exclusive events in an experiment,  then $P( E \cup F) = P( E) + P( F)$. We desire to compute the probability that $E$ occurs before $F$ , which we will denote by $p$. To compute $p$ we condition on the three mutually exclusive events $E$, $F$ , or $(E \cup F )^c$. This last event are all the outcomes not in $E$ or $F$. Letting the event $A$ be the event that $E$ occurs before $F$, we have that $p = P( A|E) P( E) + P( A|F) P(F ) + P( A|(E \cup F )^c) P( (E \cup F )^c)$ $P( A|E) = 1$ $P( A|F) = 0$ $P( A|(E \cup F)^c) = p$ since if neither $E$ or $F$ happen the next experiment will have $E$ before $F$ (and thus event $A$ with probability $p$). Thus we have that $p = P( E) + p P( (E \cup F)^c)$ $= P( E) + p (1 − P( E \cup F))$ $= P( E) + p (1 − P( E) − P( F))$ Solving for $p$ gives $p = \frac{ P( E)}{ P( E) + P( F)}$ as we were to show.  Specifically his statement  since if neither $E$ or $F$ happen the next experiment will have $E$ before $F$ (and thus event $A$ with probability $p$)
23375
I am just learning probability in my Discrete Structures class and am very lost. This is the example given in the book and I have no idea how to solve this problem.   Problem: Suppose one in 1000 people have a certain disease. Suppose medical testing is not perfect (as is the case in real life) and consequently, only 99% of the people with the disease are tested positive. Suppose 2% of the people who don’t have the disease also test positive. What is the probability of actually having the disease, given someone tests positive?  Things I do know:   People with disease = 1/1000 Tested positive with disease = 99/100 People tested positive who don't have disease = 2/100   I'm not sure where to got from here with the information that I know. What are the next steps?
112839
If I have two independent Poisson processes, X and Y, with X having a lambda 2 and Y having a lambda 3. Given that the starting value for Y is 3 and X is zero, then how do I calculate the probability of X being larger than Y at any given point, even if only for one instant in time. I realize that I can just simulate my way out of it, but that does not really seem like an elegant solution. Any help/directions would be much appreciated.  Besides the solution below, is it possible to use the CDF and pmf of a Poisson distribution instead of only its PMF, see below.  $1-\prod_{t=1}^{\infty}\bigg[\prod_{k=0}^{\infty} \bigg [1-\bigg(1-e^{-\lambda_1*t}\cdot \sum_{i=0}^{k+D+1}\bigg[\frac{(\lambda_1*t)^{i}}{i!}\bigg]\bigg)\cdot \frac{(\lambda_2*t)^{k}*e^{-\lambda_2*t}}{k!}\bigg]\bigg] $  My idea was that we find the probability that Poisson distribution X with Lambda 2 receives k+3+1 more occurrences conditioned on the fact that Poisson distribution Y gets equal to k occurrences. This is done for all k and then subsequently we do it for all points in time. Is this also a correct way of doing it?
69430
I am asked to give an example of three Poisson-distributed random variables which are pairwise independent, but are not mutually independent.  I thought of the example about the Intersection where cars are coming from one side and they can go to one of three directions left, right or keep straight, with probabilities $p \ $, $q \ $ and $1-p-q$, respectively.  However, I find it hard to prove.
83989
The probability $P(X=n)$ that an event X takes place $n$ times in a fixed period of time follows the Poisson distribution with parameter $\lambda$ i.e.  $$ P(X = n) = e^{-\lambda} \frac{\lambda ^ n}{n!}$$  I have to evaluate the probability that the event $X$ takes place an even number of times. I know that:   $$ P(X \text{ is even} ) = e^{-\lambda} \cdot \sum_{k=0}^{+\infty} \frac{\lambda^{2k}}{(2k)!}$$  but I can't solve the series.  I guess that I have to use the fact that $e^\lambda = \sum_{n = 0}^{+\infty} \lambda^n/n!$, but I got stuck.  How can I evaluate $P(X \text{ is even})$ (alternative solutions appreciated).
101013
In an urn there are $n$ red balls and $m$ blue balls. I extract them without replacement. Let $X$=time of first blue. What is $E(X)$?  I found PMF of $X$ and it is, if $k &gt; n+1$  $$P(X=k) =\frac{n(n-1) \dots (n-k+1)m}{(m+n)(m+n-1) \dots (n+m-k+1)}$$  How could I evaluate $E(X)$?  Edit: I'm looking for an explicit form of $E(X)$
13506
Given that I have a fair coin which I toss three times, how do I:  a) Find three events $A$ , $B$, and $C$ such that no two are independent but $P( A  \cap B \cap C) = P( A ) P( B) P( C)$ ?  b) Find three events $A$ , $B$, and $C$ such that every two are independent but all three are not independent: $P( A  \cap B \cap C) \not= P( A ) P( B) P( C)$ ?  How would I approach a problem like this ? I started by drawing a binary tree that ended up with 8 leaves but I have no idea how to use it to get the answers. Any help would be appreciated.
112185
We have ($P$ is probability): $P(A \cup B \cup C) = 1$ ; $P(B) = 2P(A) $ ; $P(C) = 3P(A) $ and $P(A \cap B) = P(A \cap C) = P(B \cap C) $. Prove that $P(A) \le \frac{1}{4} $. Well, I tried with the fact that $ 1 = P(A \cup B \cup C) = 6P(A) - 3P(A \cap B) + P(A \cap B \cap C) $ but I got stuck... Could anyone help me, please?
6689
A pedestrian wishes to go across a single-lane road where the cars arrive according to a Poisson process with the rate λ. The time needed for him to cross the road safely is denoted as k. He will have to wait until he sees a gap of at least k between the coming cars. If the gap between the car that just arrived and the next one is at least k, he begins to cross the road. Let T denote the random time he needs to wait by the road. What is the expected time the pedestrian needs to cross this particular road?  I've encountered several similar problems on here, but none of them gives a detailed, step-by-step explanation, so I am really confused. I would appreciate any help! Thank you in advance.
102607
Suppose a basketball player has an 80% chance of making a free throw. The player was fouled and now has two free throws. If a free throw is made it counts as one point. Let X be the number of points from the first free throw and Y be the number of points from the second free throw. Figure out the joint probabilities from X and Y (I've already done this step), the expected number of points from the two free throws and the variance for this number of points if:   a) Each free throw is an independent event   b) P(Y=1|X=1)=.9 and P(Y=1|X=0)=.4 so that making the first free throw raises the prob. a making the second free throw.   All I really need help on is the expected number of points and the variance for that number of points.
106825
Each event has two possible outcomes: a and b.  Each event is independent.  Outcome a has $X\%$ chance of happening.  If we run $Y$ number of such events, what are the chances of getting $K$ outcomes a in a row, at least once?  To put it in practical terms imagine a series of coin tosses. Each coin toss has two possible outcomes: a = heads and b = tails. Outcome a = heads has a chance of $50\%$. If we run $Y = 100$ number of coin tosses, what are the chances of getting $K = 10$ outcomes a=heads in a row, at least once?   I'm using the practical example only so I can express the idea more clearly. I'm interested in finding out the general formulas and computations so that I can run the calculations for any $X$, $Y$, or $K$.  I would also be very interested in knowing how the probabilities would change if the events would not be independent. What if the more outcomes a we have in a row the bigger the chance of getting an outcome a in the next event? What if it's the other way around, and the more outcomes a in a row we have the lower the chance of getting another a?
67700
Let $\{X_n, n \geq 1\}$ be iid Bernoulli random variables with $\Pr(X_1 = 1) = p = 1 - \Pr(X_1 = 0)$ and let $S_n = \sum\limits_{i=1}^n X_i$ be the number of successes in $n$ trials. Show $S_n$ has a binomial distribution by the following methods:  1.) Prove for $n\geq0,$ $1\leq{k}\leq{n+1}$  $$\Pr(S_{n+1} = k) = p\Pr(S_n = k - 1) + q\Pr(S_n = k)\text{ where }q = 1-p.$$  2.) Solve the recursion using generating functions.  For the first method, I think I need to use convolution of $n + 1$ Bernoulli rvs, but I cannot immediately see the progression. For the second method, I don't exactly understand how to proceed either. This is problem 1.2 from Resnick's Adventures in Stochastic Processes.
117906
$5$ tourists plan to attend Octoberfest. Each attends a location at random from the choices: Alpine, Bingeman, Concordia, Kitchener, Queens, Schwaben, Transylvania. Q What is the probability that two attend one location and three attend another same location?  I get that there are $\binom{5}{2}$ possibilities to create 2 groups. There are $\binom{7}{2}$ ways to choose 2 places so shouldn't it be $\frac{\binom{7}{2}\binom{5}{2}}{7^5}$?