November 2008 

January 2009 
bash export CVSROOT=:pserver:anonymous@cmscvs.cern.ch:/cvs_server/repositories/CMSSW export CVS_RSH=ssh cvs d `echo $CVSROOT  awk F@ '{print $1":98passwd\@"$2}'` login cvs co r DBS_2_0_2 DBS/Clients/Python cd DBS/Clients/Python source setup.shSTEP 3: Setup env/config variables
cd DBSAPI emacs dbs.config & ==> Change DBS VERSION and host URL ==> The following combinations seems to work fine: VERSION=DBS_2_0_2 URL=http://cmssrv48.fnal.gov:8383/DBS/servlet/DBSServlet [rather old] VERSION=DBS_1_0_9 URL=http://cmssrv46.fnal.gov:8080/DBS111LFNOPT/servlet/DBSServletSTEP 3: Setup env/config variables
cd UserExamples ==> Need to modify "dbsInsertEverything.py". It will better to copy directly from my (working) sample file: /uscms/home/kalanand/files_PublishToDB/dbsInsertEverything.pyMake sure to do the following changes:\\ 1. The field "primary = DbsPrimaryDataset (Name = ...., Type = ....)" defines the primary name of the dataset. The conventional naming scheme is: primary_name/process_name/type e.g., /Madgraph_ttbarjets_FNAL/CMSSW_1_6_12_Fastsim/RECO
python dbsInsertEverything.py ==> check if the dataset got into the DBS or not: dbs search query="find dataset where dataset = /test1_Madgraph_ttbarjets_FNAL/CMSSW_1_6_12_Fastsim/RECO"The usual options are: find dataset, find block, find files
dbs lsd path=/test1_Madgraph_ttbarjets_FNAL/CMSSW_1_6_12_Fastsim/RECO dbs lsb path=/test1_Madgraph_ttbarjets_FNAL/CMSSW_1_6_12_Fastsim/RECO dbs lsf path=/test1_Madgraph_ttbarjets_FNAL/CMSSW_1_6_12_Fastsim/RECOSTEP 5: Test: transfer dataset from one local server to another
python dbsMigrateWithParents.pyFor example:
python dbsMigrateWithParents.py http://cmssrv48.fnal.gov:8383/DBS/servlet/DBSServlet http://cmssrv46.fnal.gov:8080/DBS111LFNOPT/servlet/DBSServlet /test1_Madgraph_ttbarjets_FNAL/CMSSW_1_6_12_Fastsim/RECOSTEP 6: Get proxy and FINALLY publish in global DBS
vomsproxyinit voms cms python dbsMigrateWithParents.py < host_server> < destination> < dataset>For example:
python dbsMigrateWithParents.py http://cmssrv46.fnal.gov:8080/DBS111LFNOPT/servlet/DBSServlet https://cmsdbsprod.cern.ch:8443/cms_dbs_prod_global_writer/servlet/DBSServlet /Madgraph_ttbarjets_FNAL/CMSSW_1_6_12_Fastsim/RECO
#!/bin/tcsh # copy data files from FNAL dCache resilient to local area cd /pnfs/cms/WAX/resilient/kalanand/ZeeJet_OctX_7TeV/ foreach datafile ( `ls *root` ) echo "submitting: dccp $datafile /uscms/home/kalanand/cms/jet/CMSSW_3_1_4/src/JetMETCorrections/ZJet/test/$datafile" dccp $datafile /uscms/home/kalanand/cms/jet/CMSSW_3_1_4/src/JetMETCorrections/ZJet/test/$datafile end echo ""
Remember: The mean (u) and standard deviation (s) for a binomial distribution are np and √(npq), respectively.
Probability throwing dice  theory:
Throwing dice is more complicated than tossing coins, as there are more than 2 values. If you throw a single dice, then it can fall six ways, each of which is equally likely if the dice is true. So the probability of getting one particular value is 1/6. If you want either of two values it is 2/6 or 1/3, and so on.
When rolling two dice, distinguish between them in some way: a first one and second one, a left and a right, a red and a green, etc. Let (a,b) denote a possible outcome of rolling the two die, with a the number on the top of the first die and b the number on the top of the second die. Note that each of a and b can be any of the integers from 1 through 6. Here is a listing of all the joint possibilities for (a,b):
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Q: Two fair sixsided dice are rolled. What is the probability that the numbers on the dice are different?
A: 1 6/36 = 5/6
Probability of two dice totals:
Total on dice Pairs of dice Probability 2 1+1 1/36 = 3% 3 1+2, 2+1 2/36 = 6% 4 1+3, 2+2, 3+1 3/36 = 8% 5 1+4, 2+3, 3+2, 4+1 4/36 = 11% 6 1+5, 2+4, 3+3, 4+2, 5+1 5/36 = 14% 7 1+6, 2+5, 3+4, 4+3, 5+2, 6+1 6/36 = 17% 8 2+6, 3+5, 4+4, 5+3, 6+2 5/36 = 14% 9 3+6, 4+5, 5+4, 6+3 4/36 = 11% 10 4+6, 5+5, 6+4 3/36 = 8% 11 5+6, 6+5 2/36 = 6% 12 6+6 1/36 = 3%An easy way to remember the numerator in this chart:
Q: A pair of dice are rolled, find the probability that the sum is greater than 9.
A: 3/36 + 2/36 + 1/36 = 1/6
Q: A pair of dice are rolled. Find the probability that the sum is a multiple of 4.
A: 3/36 + 5/36 + 1/36 = 1/4
There is an interesting phenomenon that occurs when dealing with problems like above. If you are looking for the probability that the sum is divisible by a certain number less than 7, then you get:
If the sum is divisible by: 2 > the probability is 1/2. 3 > the probability is 1/3. 4 > the probability is 1/4. *5 > the probability is 7/36. 6 > the probability is 1/6.Note: 5 is the only abnormal one, making this easy to memorize.
Q: Conditional probability related to two dice throws. A pair if dice is thrown. If it is known that one dice shows a 4, what probability that a) the other dice shows a 5 b) the total of both the dice is greater than 7 ?
A: (a) 2/11 (b) Allowed combination: 44, 45, 46, 54, 64. Hence 5 /11.
Q: A pair if dice is thrown. One of them shows 6. What is the probability of other being 6 ?
A: 1/11
More nerdy questions from mathforum.org:
A fair die is thrown n times. The probability that there is an even number of sixes is
1/2 * [1+(2/3)^n]
Note: 0 is considered an even number.
Q: Tossing a Coin and Rolling a Die: If you toss a coin and roll a dice, what is the probability of
obtaining: a) heads and a five b) heads or a five c) tails or a two?
A: (a) 1/2 * 1/6 = 1/12 (b) 1  (11/2)(11/6) = 7/12 (c) 7/12
Q: The house rolls 10 dice, one die at a time. What are the odds of the following:
a.) A player rolling the same numbers in the same order as the house.
b.) A player rolling the same numbers as the house without regard to the order in which they were rolled.
c.) A player rolling 9 of 10 numbers the same as the house without regard to the order in which they were rolled.
A: (a) 1/6^10
(b) This is equivalent to having 6 cells (representing the possible number on the face of any dice) and 10 balls to be distributed into these 6 cells. The number of balls in any cell will represent the number of times that the cell number occurred when the 10 dice are rolled. It is possible that some cells (from 0 to 5) could be empty. So, the total number of arrangements is: C(15, 10) = 15! / (10! * 5! ) = 3003. Therefore, the desired probability = 1/3003.
(c) [N(0 empty cells) + N(1 empty cells) + … + N(5 empty cells)] / 3003 = (30 + 25 + .. + 5) / 3003 = 105 / 3003 = 5 / 143
Probability questions:
Q: Mr. Smith has two children. You already know that one of them is a girl; what's the probability that Mr. Smith has one boy and one girl?
Ans: 2/3 (possible outcomes: BG, GB, GG)
Q: A bakery makes a batch of 200 cookies in which 2000 chocolate chips were used. What is the probability that a cookie picked at random from the batch will contain at least 13 chocolate chips?
A: Probability of n chips = e^(x) * (x^n) / n! , where x = average number of chips = 10.
So, the probability = e^(10) * [10^13/ 13! + 10^14/ 14! + …]
Q: Suppose that you toss a balanced coin 100 times. What is the approximate probability that you observe less than or equal to 40 heads?
A: Sum_{k=0 to 40} C(n,k) * p^k * q^(nk) = (1/2)^100 * Sum C(100, k) = 0.03 approx.
Q: Five cards are drawn without replacement from a standard deck of playing cards. Next, five cards are drawn without replacement from a second standard deck of playing cards. And finally, five cards are drawn without replacement from a third standard deck of playing cards.
What is the probability that the Ace of Spades will appear once? Twice? Three times?
A: The chance that the Ace of Spades is chosen when five cards are dawn from one pack is: (1C1 * 51C4) / (52C5) = 0.09615
Chance that it is not selected is 1  0.09615 = 0.903846
So we have three trials, in each of which there is a probability of success of 0.096, and a chance of failure of 0.904 This is a binomial probability problem, and we have:
Probability of one success = 3C1 * 0.096 * 0.904^2 = 0.24
Probability of two successes = 3C2 * 0.096^2 * 0.904 = 0.025
Probability of three successes = 3C3 * 0.096^3 = 0.001
Q: Suppose used car salesmen tell the truth 2/5 of the time, and 1/3 of the trees in a forest are oak. If 4 used car salesmen say that a certain tree in the forest is oak, what is the probability that the tree is indeed oak?
A: Oak Prob = 1/3: (1/3).C(4,4)(2/5)^4 = (1/3)*16/625
Not oak Prob = 2/3: (2/3).C(4,0)(3/5)^4 = (2/3)*81/625
Therefore, Oak / total = 16/ (16 + 2*81) = 0.09
Q: Bob and Kyle each have $18. They flip a fair coin repeatedly. If the coin comes up tails, Bob pays Kyle $1. If the coin comes up heads, Kyle pays Bob $1. The game is finished when either one of them has no money left. What is the expected number of coin flips this game will last?
A: Let's assume that the players start with nothing, and the game lasts until one of the players has won either S or +S dollars (in this case, S = 18). Let's say x[n] represents the expected duration of the game when the player has won n dollars. We are trying to compute x[0].
x[n] = S^2  n^2
So, x[0] = 18^2 = 324 coin flips.
Q: Sue makes 70% of the free throws she attempts. She shoots three free throws in her warmup before a game. What is the probability that Sue makes two or more of the three free throws?
A: 3C2 * (0.70)^2 * (0.30)^1 + 3C3 * (0.70)^3 = 0.441 + 0.343 = 0.784
Q: There are seven different car keys in a box, including mine. I'm going to randomly remove one key at a time, and then try to start my car. If the key doesn't start the car, I'll discard the key. We know that the probability is one in seven that I'll retrieve my own car key on the first try. But what is the probability that I'll get my key on the second try?
A: It's still 1/7. The probability that the second key is your key contains two elements:
The first key is not your key AND the second key is. => (11/7) * (1/6) = 1/7
Q: Law of Large Numbers and the Gambler's Fallacy. Should you get the same total, on average, when you make three throws of three dice each as when you throw nine dice at once?
A: The sum, the average, and the probabilities of getting a particular value on one die or more aren't affected by whether the dice are rolled one at a time, in groups of 3, or all 9 at once. These are called "independent events," and the order in which they happen doesn't affect the outcome.
Q: If a car has 2 headlights with an average lifespan of 2500 hours,whose probability of failure can be modeled by: f(t) = 1/u e ^(t/u) for the mean value u = 2500, what is the probability that both lamps will fail within 2500 hours.
A: The probability that either will fail within 2500 hours is 0.632 (radioactivity). Assuming that multiple failures would be independent events, the probability that BOTH fail within 2500 hours = 0.632^2 = 0.40
Q: What is the minimum number of people we would need to assemble in a group such that the probability that at least one person in the group has the same birthday as you is greater than 50%?
A: Find minimum n such that (364/365)^n < 0.5 => 253 people !!!!!
Q: Having thrown a die 10 times, what is the probability of rolling all six numbers? Order does not matter.
A: T(10,6)/6^10 where T(10,6) is the coefficient of x^10 in the expansion [e^x  1]^6. Plugging in all the numbers the probability is 0.27 !!
Q: What is the probability of having a Social Security Number comprised of only two digits (say 1s and 2s) ?
A: 10C2 * (2^9  2)/ 10^9 = 22950/ 10^9 ~ 1/44000. Note that I needed to subtract 2 from 2^9 because I need to exclude numbers made exclusively using a single digit.
Q: If you flip a coin 10 times, what is the probability of getting at least 4 heads?
A: [10C4 + 10C5 + 10C6 + 10C7+10C8 + 10C9 + 10C10] (1/2)^10
= 1  [1+10C1 + 10C2 + 10C3]/2^10 = 1  176/1024 = 82.8%
Q: In a box there are nine fair coins and one twoheaded coin. One coin is chosen at random and tossed twice. Given that heads show both times, what is the probability that the coin is the twoheaded one? What if it comes up heads for three tosses in a row?
A: Before performing the experiment, here are the three possible outcomes
flip two heads (1/4) / choose fair coin (9/10) / \flip anything else (3/4) 10 coins \ choose twoheaded coin (1/10) > flip 2 heads (1/1)The top row has probability (9/10)*(1/4) = 9/40
Q: How can you use a 6sided Die to Generate a Random Number from 1 to 7 ?
Ans:Roll the dice twice, keeping track of the first and second roll. There are 36 outcomes, let's discard 66. If you get a (6,6), just reroll the die twice again until you get a non(6,6). Now there are 35 equallylikely outcomes, so divide them into 7 groups of 5 corresponding to the 7 choices among which you want to choose.
Q: Of the nine members of the board of trustees of a college, five agree with the president on a certain issue. The president selects three trustees at random and asks their opinions. What is the probability that at least two of them will agree with him?
A: (5C3 + 5C2 * 4) / 9C3 = (10 + 4 * 10) / 84 = 50/84 = 25/42 ~ 60%
Q: Suppose there are two full bowls of cookies. Bowl #1 has 10 chocolate chip and 30 plain cookies, while bowl #2 has 20 of each. Our friend Fred picks a bowl at random, and then picks a cookie at random. We may assume there is no reason to believe Fred treats one bowl differently from another, likewise for the cookies. The cookie turns out to be a plain one. How probable is it that Fred picked it out of bowl #1?
A: 30/(30+20) = 0.6. Note that the probability was 0.5 before Fred picked up the cookie.
#!/bin/tcsh setenv DIR 00_09 cd /uscmst1b_scratch/lpc1/3DayLifetime/kalanand/$DIR/ foreach datafile ( `ls *` ) echo "submitting: file://localhost//uscmst1b_scratch/lpc1/3DayLifetime/kalanand/$DIR/$datafile srm://cmssrm.fnal.gov:8443/srm/managerv2?SFN=/11/store/user/kalanand/$DIR/$datafile" srmcp 2 debug=true "file://localhost//uscmst1b_scratch/lpc1/3DayLifetime/kalanand/$DIR/$datafile" "srm://cmssrm.fnal.gov:8443/srm/managerv2?SFN=/11/store/user/kalanand/$DIR/$datafile" end echo ""