Math 141. Lecture 3: The Binomial Distribution. Albyn Jones 1. 1 Library jones/courses/141


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1 Math 141 Lecture 3: The Binomial Distribution Albyn Jones 1 1 Library jones/courses/141
2 Outline Coin Tossing Coin Tosses
3 Independent Coin Tosses Crucial Features Dichotomous Trials: Each toss results in either Heads, H, or Tails T. Independence: Successive tosses are independent; knowing we got H on the first toss does not help us predict the outcome of the second toss. Constant probability: Each trial has the same probability P(H) = 1/2 = P(T ) (a fair coin ).
4 Examples of Different Experiments Binomial: Count the number of Heads in a fixed number of tosses. Geometric: Count the number of Tails before the first Head. Negative Binomial: Count the number of Tails before before the kth Head.
5 Random Variables Random Variable: A random variable is a function mapping points in the sample space to R, the real numbers. Example: Let X be the number of Heads in 3 independent tosses of a fair coin.
6 A Binomial Random Variable Toss a fair coin 3 times, count the number of Heads. Let X be the number of Heads. What are the possible outcomes? {HHH} X = 3 {HHT } {HTH} {THH} X = 2 {HTT } {THT } {TTH} X = 1 {TTT } X = 0
7 Probabilities Coin Tossing Thus we can compute probabilities for the random variable (RV) X is we can compute probabilities of events in the original sample space. P(X = 3) = P({HHH}) P(X = 2) = P({HHT } {HTH} {THH}) P(X = 1) = P({HTT } {THT } {TTH}) P(X = 0) = P({TTT })
8 Probabilities Coin Tossing Successive tosses are independent, so and P(X = 3) = P({HHH}) = P(H)P(H)P(H) = P(X = 0) = P({TTT }) = P(T )P(T )P(T ) = ( ) ( ) In fact, the probability of any sequence of 3 tosses is the same, for example P({HHT }) = P(H)P(H)P(T ) = ( ) 1 3 2
9 More Probabilities Since probabilities of unions of disjoint events add, we just have to count the number of sequences of three tosses with 2 Heads to get P(X = 2): P(X = 2) = P({HHT } {HTH} {THH}) = P({HHT }) + P({HTH}) + P({THH}) Again, due to independence, the order of getting the two Heads and one Tail doesn t matter, so ( ) 1 3 P({HHT }) = P({HTH}) = P({THH}) = 2 Thus P(X = 2) = 3 ( ) 1 3 = 3 2 8
10 Finally: Probabilities for 3 tosses ( ) 1 3 P(X = 3) = 1 = Note: ( ) 1 3 P(X = 2) = 3 = ( ) 1 3 P(X = 1) = 3 = ( ) 1 3 P(X = 0) = 1 = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1
11 X Binomial(n, 1/2) Count Heads in n independent tosses of a fair coin As with 3 independent tosses, any sequence of n independent tosses of a fair coin has the same probability. Example: for 8 tosses, P(HHHHTTTT ) = P(HTHTHTHT ) = P(TTTTTTHH) = ( ) The point: to compute P(X = k) we have to count the number of sequences of n symbols {H, T } with k H s, and thus (n k) T s, then multiply by the probability of any sequence of n tosses.
12 Counting Things: Permutations How many ways are there to rearrange the numbers 1, 2, 3, 4? We could list them all, with some effort: {1, 2, 3, 4}, {2, 1, 3, 4}, {3, 1, 2, 4}, {4, 1, 2, 3},... We could be clever: There are 4 choices for the first position. After choosing the first, there remain 3 choices for the second position. After choosing the first and second positions, there are 2 possibilities for the third, and only 1 for the fourth = 4! = 24
13 Permutations of n objects How many ways are there to rearrange the numbers 1, 2, 3,..., n? There are n choices for the first position. After choosing the first, there remain (n 1) choices for the second position. After choosing the first and second positions, there are (n 2) possibilities for the third, and so forth. n (n 1) (n 2)... 1 = n! n! gets big quickly: 10! =
14 Example: Permutations of 4 objects Each row lists permuations beginning with the same object, grouped by the choices for the second object:
15 Practice with Factorials What is 1!? What is 2!? What is 3!? What is 0!?
16 A Formal definition for Factorials The standard recursive definition is For n > 0 Z, 0! = 1 n! = n (n 1)! It is useful to remember this recursive definition! For example it makes obvious the relation n! n = (n 1)!
17 Combinations Coin Tossing How many ways are there to rearrange the 5 symbols H, H, H, T, T, that is three Heads and two Tails? Label them uniquely: H 1, H 2, H 3, T 4, T 5 We have 5 objects, there are 5! orders. Some of those orders are redundant: {H 1, H 2, H 3, T 4, T 5 } is indistinguishable from {H 3, H 2, H 1, T 5, T 4 } or {H 3, H 1, H 2, T 4, T 5 }. Divide out the number of indistinguishable (or equivalent) patterns of 3 Heads, i.e. 3!, and 2 Tails, i.e. 2!. Answer: 5! 3! 2! = 5 4 3! = 5 4 3! 2! 2 1 = 10
18 Example: 4 Coin Tosses 4 HHHH 3 HHHT, HHTH, HTHH, THHH ( 4 4) ( 4 3) 2 HHTT, HTHT, THHT, TTHH, THTH, HTTH ( ) TTTH, TTHT, THTT, HTTT 0 TTTT ( 4 1) ( 4 0)
19 Sequences of k Heads in n tosses For 3 H and 2 T the number of possible orders was 5! 3! 2! For k H and (n k) T in n tosses the number of possible orders will be n choose k : the number of ways to select k objects out of a set of size n. ( ) n k Note: ( ) n = n k = n! k! (n k)! n! (n k)! k! = ( ) n k
20 Practice with Binomial Coefficients What is each of the following? ( ) n 0 ( ) n n ( ) n 1 ( n ) n 1 ( ) n 2
21 In general, we might be working with dichotomous trials where the event of interest has probability p, possibly not 1/2. Most of what we have learned about cointossing carries over to the general case.
22 The Binomial Distribution X Binomial(n, p) Dichotomous Trials: Each trial results in either a Success, S, or a Failure F. Independence: Trials are mutually independent; knowing we got S on one trial does not help us predict the outcome of any other trial. Constant probability: Each trial has the same probability P(S) = p, and P(F) = 1 p. X counts the number of S s. n the number of trials is fixed.
23 Binomial(n, p) Probabilities Independence: Every sequence of n trials with k successes has the same probability! P(SSF ) = P(S)P(S)P(F ) = P(F )P(S)P(S) = P(FSS) Let p = P(S) and q = 1 p = P(F), then for a RV X Binomial(n, p) ( ) n P(X = k) = p k q n k k Again, ( n k) counts the number of sequences of length n with k successes.
24 Example Coin Tossing Suppose we roll a fair die 5 times. What is the probability we get no ones? What Binomial distribution should we use? Let S be the event we roll a 1. If the die is fair, p = P(S) = 1/6. n = 5, as there are 5 trials. Let X be the number of 1 s we get in the 5 rolls. Then Thus P(X = 0) = ( 5 0 X Binomial(5, 1 6 ) ) ( 1 6 ) 0 ( ) 5 5 =
25 Binomial(10,1/2)
26 Binomial(10,1/5)
27 Sums of Binomial RV s Suppose that X Binomial(n, p) and Y Binomial(m, p) are independent Binomial RV s with the same probability p. What is the distribution of X + Y? X + Y Binomial(n + m, p)!
28 R functions Coin Tossing Density Function: P(X = k) = dbinom(k, n, p) (Cumulative) Distribution Function: P(X k) = pbinom(k, n, p) Random numbers: rbinom(n, n, p) Last Example: X Binomial(5, 1 6 ), P(X = 0) = dbinom(0, 5, 1/6) =
29 Binomial Random Walk Random Walk functions on 141 website RW = function(n) { x = sample(c(1,1),size=n,replace=t) rw = cumsum(x) plot(1:n,rw,xlab="n",ylim=c(3.1*sqrt(n),3.1*sqr abline(h=0,lty=2) } RW1= function(n,color="blue") { x = sample(c(1,1),size=n,replace=t) rw = cumsum(x) lines(1:n,rw,col=color) }
30 Summary Coin Tossing There are n! permutations of n objects. Binomial coefficients, the number of subsets of size k from a set of n objects: ( ) n n! = k k! (n k)! The Binomial Distribution: P(X = k) = ( ) n p k q n k k
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