The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events Permutations 4 CDs fit the CD wallet.

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The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events Permutations 4 CDs fit the CD wallet you carry to school each day. You can arrange them in the holder by release date, alphabetically by artist or title, or ranked in order of preference. For simplicity label them ABCD and try writing out an exhaustive listing of all possible arrangements: How many distinct arrangements are possible? ABCD BACD CBAD DBCA ACBD ABDC BCDA ACDB BADC BCAD CBDA BDCA CDAB CDBA ADCB CADB BDAC DCAB DABC DBAC DCBA ADBC DACB CABD If none have been repeated or missed, 24 different arrangements are possible.

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events By the time you have made that choice, there is only one left to be last. That makes for (4 possible 1st place holders)  (3 possible remaining 2nd place holders)  (2 ways to fill the next-to-the-last slot)  (1 last-place-holder) = 4·3·2·1=24. There are 4 titles to choose from as 1st in the holder... 4 different ways to start the arrangement. That leaves 3 from which to select the 2nd... then two to pick from to hold the next to last slot.

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events Each arrangement is called a permutation, and we employ the special notation of N! to represent the string of factors counting down from N to 1: 4! (read "four factorial") = 4·3·2·1=24 or 6! = 6·5·4·3·2·1=720 N! = N (N - 1) (N - 2) …3 · 2 · 1

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events Your group of 10 finds 10 seats together in a front row of the theater! How many different seating arrangements are possible? Express using factorials the number of ways a deck of 52 cards can be shuffled. 10! = 10  9  8  7  6  5  4  3  2  1 = 3,628,800 52! =  10 67

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events Combinations Four cards are set out on the table. You are asked to pick any three. How many different choices are available to you? In other words: How many different (sub)sets of 3 can you build out of a pool of 4 objects? “Picking 3” in this case is the same as “rejecting 1” (deciding which one NOT to pick) there are obviously 4 ways to do this. Listing them is easy: ABCABDADCBCD. We say 4 C 3 = 4 (the number of different combinations of 3 taken from a total of 4 is equal to 4).

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events We can count off other combinations: 4 C 2 = AB AC AD BC BD CD 4 C 1 = A B C D 6 4 Just check out: 4 C 3 = 4C2=4C2=

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events In a club with 10 members, how many ways can a committee of 3 be selected? 5 cards are drawn (one at a time) from a well-shuffled deck. How many different hands are possible? 10 C 3 = 52 C 5 =

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events A “fair” coin is flipped at the start of a football game to determine which team receives the ball. The “probability” that the coin comes up HEAD s is expressed as A. 50/50“fifty-fifty” B. 1/2 “one-half” C. 1:1 “one-to-one” In betting parlance the odds are 1:1; we say the chances are 50/50, but the mathematical probability is ½.

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events There are 36 possible outcomes for the toss of 2 six-sided dice. If each is equally likely, the most probable total score of any single roll is? A. 4 B. 5 C. 6 D. 7 E. 8 F. 9

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events A green and red die are rolled together. What is the probability of scoring an 11? A. 1/4B. 1/6C. 1/8 D. 1/12E. 1/18F. 1/36

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events “Snake eyes” give the minimum roll. “Boxcars” give the maximum roll. The probability of rolling any even number, Probability(even) ______ Probability(odd), the probability of rolling any odd number. A. > B. =C. <

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events Die Total Number of ways to score die totals

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events A coin is tossed twice in succession. The probability of observing two heads (HH) is expressed as A. 1/2B. 1/4 C. 1D. 0

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events A coin is tossed twice in succession. The probability of observing two heads (HH) is expressed as A. 1/2B. 1/4 C. 1D. 0 It is equally likely to observe two heads (HH) as two tails (TT) T) True.F) False.

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events A coin is tossed twice in succession. The probability of observing two heads (HH) is expressed as A. 1/2B. 1/4 C. 1D. 0 It is equally likely to observe two heads (HH) as two tails (TT) T) True.F) False. It is equally likely for the two outcomes to be identical as to be different. T) True.F) False.

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events A coin is tossed twice in succession. The probability of observing two heads (HH) is expressed as A. 1/2B. 1/4 C. 1D. 0 It is equally likely to observe two heads (HH) as two tails (TT) T) True.F) False. It is equally likely for the two outcomes to be identical as to be different. T) True.F) False. The probability of at least one head is A. 1/2B. 1/4 C. 3/4D. 1/3

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events The probability of “throwing a 7” P (7) = 6/36 = 1/6 P (7±5) = ?

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events P (7) = 6/36 = 1/6 P (7±5) = 36/36 =1 P (7±1) = P (7±2) = the full range!

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events P (7) = 6/36 = 1/6 P (7±5) = 36/36 =1 P (7±1) =16/36 = 4/9 P (7±2) = 24/36 = 2/3 the full range!

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events If events occur randomly in time, the probability that the next event occurs in the very next second is as likely as it not occurring until 10 seconds from now. T) True. F) False.

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events P(1)  Probability of the first count occurring in in 1st second P(10)  Probability of the first count occurring in in 10th second i.e., it won’t happen until the 10th second ??? P(1) = P(10) ??? = P(100) ??? = P(1000) ??? = P(10000) ???

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events Imagine flipping a coin until you get a head. Is the probability of needing to flip just once the same as the probability of needing to flip 10 times? Probability of a head on your 1st try, P(1) = Probability of 1st head on your 2nd try, P(2) = Probability of 1st head on your 3rd try, P(3) =

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events Probability of a head on your 1st try, P(1) =1/2 Probability of 1st head on your 2nd try, P(2) =1/4 Probability of 1st head on your 3rd try, P(3) =1/8 Probability of 1st head on your 10th try, P(10) =

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events What is the total probability of ALL OCCURRENCES? P(1) + P(2) + P(3) + P(4) + P(5) + =1/2+ 1/4 + 1/8 + 1/16 + 1/32 +

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events A six-sided die is rolled repeatedly until it gives a 6. What is the probability that one roll is enough?

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events A six-sided die is rolled repeatedly until it gives a 6. What is the probability that one roll is enough? 1/6 What is the probability that it will take exactly 2 rolls?

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events A six-sided die is rolled repeatedly until it gives a 6. What is the probability that one roll is enough? 1/6 What is the probability that it will take exactly 2 rolls? (probability of miss,1st try)  (probability of hit)=

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events A six-sided die is rolled repeatedly until it gives a 6. What is the probability that one roll is enough? 1/6 What is the probability that it will take exactly 2 rolls? (probability of miss, 1st try)  (probability of hit)= What is the probability that exactly 3 rolls will be needed?

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events CROP workshop participants have seen counts for RANDOM EVENTS fluctuate counting cloud chamber tracks Geiger-Meuller tubes clicking in response to a radioactive source “scaling” the cosmic ray singles rate of a detector (for a lights on/lights off response)

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events Cosmic rays form a steady background impinging on the earth equally from all directions measured rates NOT literally CONSTANT long term averages are just reliably consistent These rates ARE measurably affected by Time of day Direction of sky Weather conditions

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events You set up an experiment to observe some phenomena …and run that experiment for some (long) fixed time… but observe nothing: You count ZERO events. What does that mean? If you observe 1 event in 1 hour of running Can you conclude the phenomena has a ~1/hour rate of occurring?

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events Random events arrive independently unaffected by previous occurrences unpredictably 0 sectime  A reading of 1 could result from the lucky capture of an exceeding rare event better represented by a much lower rate (~ 0? ). or the run period could have just missed an event (starting a moment too late or ending too soon).

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events A count of 1 could represent a real average as low as 0 or as much as 2 1 ± 1 A count of 2 2 ± 1? ± 2? A count of ± at least a few? A count of ± ?

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events The probability of a single COSMIC RAY passing through a small area of a detector within a small interval of time  t can be very small: p << 1 cosmic rays arrive at a fairly stable, regular rate when averaged over long periods the rate is not constant nanosec by nanosec or even second by second this average, though, expresses the probability per unit time of a cosmic ray’s passage

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events would mean in 5 minutes we should expect to count about A.6,000 events B. 12,000 events C. 72,000 events D. 360,000 events E. 480,000 events F. 720,000 events 1200 Hz = 1200/sec Example: a measured rate of

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events would mean in 3 millisec we should expect to count about A.0 events B. 1 or 2 events C. 3 or 4 events D. about 10 events E. 100s of events F. 1,000s of events 1200 Hz = 1200/sec Example: a measured rate of 1 millisec = 10  3 second

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events would mean in 100 nanosec we should expect to count about A.0 events B. 1 or 2 events C. 3 or 4 events D. about 10 events E. 100s of events F. 1,000s of events 1200 Hz = 1200/sec Example: a measured rate of 1 nanosec = 10  9 second

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events The probability of a single COSMIC RAY passing through a small area of a detector within a small interval of time  t can be very small: p << 1 for example (even for a fairly large surface area) 72000/min=1200/sec =1200/1000 millisec =1.2/millisec = /  sec = /nsec

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events The probability of a single COSMIC RAY passing through a small area of a detector within a small interval of time  t can be very small: p << 1 The probability of NO cosmic rays passing through that area during that interval  t is A. p B. p 2 C. 2p D.( p  1) E. (  p)

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events The probability of a single COSMIC RAY passing through a small area of a detector within a small interval of time  t can be very small: p << 1 If the probability of one cosmic ray passing during a particular nanosec is P(1) = p << 1 the probability of 2 passing within the same nanosec must be A. p B. p 2 C. 2p D.( p  1) E. (  p)

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events The probability of a single COSMIC RAY passing through a small area of a detector within a small interval of time  t is p << 1 the probability that none pass in that period is ( 1  p )  1 While waiting N successive intervals (where the total time is t = N  t ) what is the probability that we observe exactly n events? × ( 1  p ) ??? ??? “misses” p n n “hits” × ( 1  p ) N-n N-n“misses”

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events While waiting N successive intervals (where the total time is t = N  t ) what is the probability that we observe exactly n events? P(n) = n C N p n ( 1  p ) N-n From the properties of logarithms ln (1  p) N-n = ln (1  p) ln x  log e x e= ??? ln (1  p) N-n = (N  n) ln (1  p)

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events ln (1  p) N-n = (N  n) ln (1  p) and since p << 1 ln (1  p)   p ln (1  p) N-n = (N  n) (  p) from the basic definition of a logarithm this means e ???? = ???? e -p(N-n) = (1  p) N-n

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events P(n) = p n ( 1  p ) N-n P(n) = p n e -p(N-n) P(n) = p n e -pN If we have to wait a large number of intervals, N, for a relatively small number of counts,n n<<N

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events P(n) = p n e -pN And since N - (n-1)  N (N) (N) … (N) = N n for n<<N

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events P(n) = p n e -pN P(n) = e -Np Consider an example…

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events P(n) = e  4 If the average rate of some random event is p = 24/min = 24/60 sec = 0.4/sec what is the probability of recording n events in 10 seconds ? P(0) = P(4) = P(1) = P(5) = P(2) = P(6) = P(3) = P(7) = e -4 =

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events P(n) = e -Np Hey! What does Np represent?

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events Another useful series we can exploit , mean = n=0 term n / n! = 1/(???)

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events , mean let m = n  1 i.e., n =      0 N )!( )N( )(N m m p m p p e what’s this?

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events , mean  = (Np) e  Np e Np  = Np

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events  = Np P(n) = e  Poisson distribution probability of finding exactly n events within time t when the events occur randomly, but at an average rate of   (events per unit time)

The Cosmic Ray Observatory Project High Energy Physics Group The University of Nebraska-Lincoln Counting Random Events  =1  =4  =8