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Lectures prepared by: Elchanan Mossel Yelena Shvets Introduction to probability Stat 134 FAll 2005 Berkeley Follows Jim Pitman’s book: Probability Section.

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Presentation on theme: "Lectures prepared by: Elchanan Mossel Yelena Shvets Introduction to probability Stat 134 FAll 2005 Berkeley Follows Jim Pitman’s book: Probability Section."— Presentation transcript:

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2 Lectures prepared by: Elchanan Mossel Yelena Shvets Introduction to probability Stat 134 FAll 2005 Berkeley Follows Jim Pitman’s book: Probability Section 2.1

3 Binomial Distribution Toss a coin 100, what’s the chance of 60 ? hidden assumptions: n-independence; probabilities are fixed.

4 Magic Hat: Each time we pull an item out of the hat it magically reappears. What’s the chance of drawing 3 I-pods in 10 trials? ={ } ( ( ))

5 Binomial Distribution Suppose we roll a die 4 times. What’s the chance of k ? Let. = { } k=0

6 Binomial Distribution Suppose we roll a die 4 times. What’s the chance of k ? Let. = { } k=1

7 Binomial Distribution Suppose we roll a die 4 times. What’s the chance of k ? Let. = { } k=2

8 Binomial Distribution Suppose we roll a die 4 times. What’s the chance of k ? Let. = { } k=3

9 Binomial Distribution Suppose we roll a die 4 times. What’s the chance of k ? Let. = { } k=4

10 Binomial Distribution Suppose we roll a die 4 times. What’s the chance of k ? Let. = { } k=0 k=1 k=2 k=3 k=4

11 Binomial Distribution How do we count the number of sequences of length 4 with 3. and 1. ?.1.1 1.1... 1.2..2. 1..... 1.. 3.. 3.. 1....... 1... 4... 6...4... 1....

12 Binomial Distribution This is the Pascal’s triangle, which gives, as you may recall the binomial coefficients.

13 Binomial Distribution 1a1a 1b1b 1a21a2 2ab2ab 1b21b2 1a31a3 3a2b3a2b 3ab23ab2 1b31b3 1a41a4 4a3b4a3b 6a2b26a2b2 4ab34ab3 1b41b4 (a + b) = (a + b) 2 = (a + b) 3 = (a + b) 4 =

14 Newton’s Binomial Theorem

15 Binomial Distribution For n independent trials each with probability p of success and (1-p) of failure we have This defines the binomial(n,p) distribution over the set of n+1 integers {0,1,…,n}.

16 Binomial Distribution This represents the chance that in n draws there was some number of successes between zero and n.

17 Example: A pair of coins A pair of coins will be tossed 5 times. Find the probability of getting on k of the tosses, k = 0 to 5. binomial(5,1/4)

18 Example: A pair of coins A pair of coins will be tossed 5 times. Find the probability of getting on k of the tosses, k = 0 to 5. # =k To fill out the distribution table we could compute 6 quantities for k = 0,1,…5 separately, or use a trick.

19 Binomial Distribution: consecutive odds ratio Consecutive odds ratio relates P(k) and P(k-1).

20 543210k Example: A pair of coins Use consecutive odds ratio to quickly fill out the distribution table for binomial (5,1/4):

21 Binomial Distribution We can use this table to find the following conditional probability: P(at least 3 | at least 1 in first 2 tosses) = P(3 or more & 1 or 2 in first 2 tosses) P(1 or 2 in first 2)

22 543210k bin (5,1/4): 10k 2 bin (2,1/4):

23 Stirling’s formula How useful is the binomial formula? Try using your calculators to compute P( 500 H in 1000 coin tosses ) directly: Your calculator may return error when computing 1000!. This number is just too big to be stored.

24 The following is called the Stirling’s approximation. It is not very useful if applied directly : n n is a very big number if n is 1000.

25 Stirling’s formula:

26 Binomial Distribution Toss coin 1000 times; P( 500 in 1000 | 250 in first 500 )= P( 500 in 1000 & 250 in first 500 )/P( 250 in first 500 )= P( 250 in first 500 & 250 in second 500 )/P( 250 in first 500 )= P( 250 in first 500 ) P( 250 in second 500 )/ P( 250 in first 500 )= P( 250 in second 500 )=

27 Binomial Distribution: Mean Question: For a fair coin with p = ½, what do we expect in 100 tosses?

28 Binomial Distribution: Mean Recall the frequency interpretation: p ¼ #H/#Trials So we expect about 50 H !

29 Binomial Distribution: Mean Expected value or Mean  of a binomial(n,p) distribution  = #Trials £ P(success) = n p slip !!

30 Binomial Distribution: Mode Question: What is the most likely number of successes? Mean seems a good guess.

31 Binomial Distribution: Mode Recall that To see whether this is the most likely number of successes we need to compare this to P(k in 100) for every other k.

32 Binomial Distribution: Mode The most likely number of successes is called the mode of a binomial distribution.

33 Binomial Distribution: Mode If we can show that for some m P(1) · … P(m-1) · P(m) ¸ P(m+1) … ¸ P(n), then m would be the mode.

34 Binomial Distribution: Mode so we can use successive odds ratio: to determine the mode.

35 Binomial Distribution: Mode successive odds ratio: If we replace · with > the implications will still hold. >> >> >> So for m= b np+p c we get that P(m-1) · P(m) > P(m+1).

36 Binomial Distribution: Mode - graph SOR k mode = 50

37 Binomial Distribution: Mode - graph distr k mode = 50

38 Lectures prepared by: Elchanan Mossel Yelena Shvets Introduction to probability Stat 134 FAll 2005 Berkeley Follows Jim Pitman’s book: Probability Section 2.1


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