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Expected Value Reprise CP Canoe Club Duck Derby  Maximum of 2 000 tickets sold  $5/ticket Prizes  1) 12 VIP tickets to Cirque du Soleil ($2,000) 

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Presentation on theme: "Expected Value Reprise CP Canoe Club Duck Derby  Maximum of 2 000 tickets sold  $5/ticket Prizes  1) 12 VIP tickets to Cirque du Soleil ($2,000) "— Presentation transcript:

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2 Expected Value Reprise CP Canoe Club Duck Derby  Maximum of 2 000 tickets sold  $5/ticket Prizes  1) 12 VIP tickets to Cirque du Soleil ($2,000)  2) 32gb BlackBerry Playbook ($600)  3) $250 Should Mr. Lieff buy a ticket?

3 Expected Value Calculation For 2 000 tickets  E(X) =1/2000 (2000) + 1/2000(600) + 1/2000(250)  = 1.425 Suppose Mr. Lieff receives a hot tip that only 1 500 tickets will be sold.  E(X) =1/1500 (2000) + 1/1500(600)+1/1500(250)  = 1.9

4 Probability Distributions and Expected Value Chapter 5.1 – Probability Distributions and Predictions MSIP / Home Learning: p. 277 #1-5, 9, 12, 13

5 Probability Distributions of a Discrete Random Variable a discrete random variable X is a variable that can take on only a finite set of values for example, rolling a die can only produce numbers in the set {1,2,3,4,5,6} rolling 2 dice can only produce numbers in the set {2,3,4,5,6,7,8,9,10,11,12} choosing a card from a standard deck (ignoring suit) can produce only the cards in the set {A,2,3,4,5,6,7,8,9,10,J,Q,K}

6 Probability Distribution the probability distribution of a random variable x, is a function which provides the probability of each possible value of x may be represented as a table of values or a graph ex, rolling a die:

7 Probability Distribution for 2 Dice

8 What would a probability distribution graph for three dice look like? We will try it! Using three dice, figure out how many outcomes there are Then find out how many possible ways there are to create each of the possible outcomes Fill in a table like the one below Now you can make the graph Outcome3456789… # ways1

9 Probability Distribution for 3 Dice Outcome345678910 # cases13610152128

10 So what does an experimental distribution look like? A simulated dice throw was done a million times using a computer program and generated the following data What is the most common outcome? Does this make sense?

11 Back to 2 Dice What is the expected value of throwing 2 dice? How could this be calculated? So the expected value of a discrete variable X is the sum of the values of X multiplied by their probabilities

12 Example 1a: tossing 3 coins What is the likelihood of at least 2 heads? It must be the total probability of tossing 2 heads and tossing 3 heads P(X = 2) + P(X = 3) = ⅜ + ⅛ = ½ so the probability is 0.5 X0 heads1 head2 heads3 heads P(X)⅛⅜⅜⅛

13 Example 1b: tossing 3 coins What is the expected number of heads? It must be the sums of the values of x multiplied by the probabilities of x 0P(X = 0) + 1P(X = 1) + 2P(X = 2) + 3P(X = 3) = 0(⅛) + 1(⅜) + 2(⅜) + 3(⅛) = 1½ So the expected number of heads is 1.5 X0 heads1 head2 heads3 heads P(X)⅛⅜⅜⅛

14 Combinations Recall that C(n, r) is the number of ways r objects can be chosen from n when order doesn’t matter.

15 Example 2a: Selecting a Committee of three people from a group of 4 men and 3 women What is the probability of having at least one woman on the team? There are C(7,3) or 35 possible teams C(4,3) = 4 have no women C(4,2) x C(3,1) = 6 x 3 = 18 have one woman C(4,1) x C(3,2) = 4 x 3 = 12 have 2 women C(3,3) = 1 has 3 women

16 Example 2a cont’d: selecting a committee What is the likelihood of at least one woman? It must be the total probability of all the cases with at least one woman P(X = 1) + P(X = 2) + P(X = 3) = 18/35 + 12/35 + 1/35 = 31/35 X0 women1 woman2 women3 women P(X)4/3518/3512/351/35

17 Example 2b: selecting a committee What is the expected number of women? 0P(X = 0) + 1P(X = 1) + 2P(X = 2) + 3P(X = 3) = 0(4/35) + 1(18/35) + 2(12/35) + 3(1/35) = 1.3 (approximately) X0 women1 woman2 women3 women P(X)4/3518/3512/351/35

18 MSIP / Home Learning p. 277 #1-5, 9, 12, 13

19 Pascal’s Triangle Chapter 5.2 – Probability Distributions and Predictions Due now: p. 277 #1-5, 9, 12, 13 MSIP/Home Learning: p. 289 #1, 2aceg, 6-8, 11-12

20 How many routes are there to the top right-hand corner? you need to move up 4 spaces and over 5 spaces This is the same as rearranging the letters NNNNEEEEE This can be calculated by C(9,4) or C(9,5) = 126 ways

21 Permutation AND Combination? As a permutation:  There are 9 moves  9! However, 4 are identical N moves and 5 are identical E moves  Divide by 4! due to identical arrangements of Ns  Divide by 5! due to identical arrangements of Es   9! ÷ 4!5! = 126 As a combination:  There are 9 moves required _ _ _ _ _ _ _ _ _  Choose 4 to be Ns, the rest are Es  C(9, 4) = 126  Choose 5 to be Es, the rest are Ns  C(9, 4) = 126

22 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Pascal’s Triangle the outer values are always 1 the inner values are determined by adding two values diagonally above

23 Pascal’s Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 sum of each row is a power of 2 1 = 2 0 2 = 2 1 4 = 2 2 8 = 2 3 16 = 2 4 32 = 2 5 64 = 2 6

24 Pascal’s Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 Uses? binomial theorem combinations!  e.g. choose 2 items from 5  go to the 5 th row, the 2 nd number = 10 (always start counting at 0) modeling the electrons in each shell of an atom (google ‘Pascal’s Triangle electron’)

25 Pascal’s Triangle – Cool Stuff 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 each diagonal is summed up in the next value below and to the left called the “hockey stick” property there may even be music hidden in it http://www.geocities.com/Vi enna/9349/pascal.mid http://www.geocities.com/Vi enna/9349/pascal.mid

26 Music: Three Parts in D Alternating Octatonic The piece of music has three parts: 1) A treble part where the center numbers 1, 2 and 6 are the tonic D of the D alternating octatonic scale. The other vertical lines of numbers represent one scale step; a descending scale step to the left and an ascending scale step to the right. All ways of reaching all the numbers in the triangle are mapped to consecutive sixteenths, starting from the top and going down south west first to every new row, inwards. This part is repeated. 2) The main treble part. It has the structure A 1 B A 2 B. It starts with the left side of the symmetric triangle, including the center (A 1 ), the numbers are the number of scale steps to ascend (south east arrows) or descend (south west arrows). After an intermission (B) where this part doubles the first part, the right side is initiated (A 2 ) and it is the mirror image of A 1. 3) A walking bass part (quarter notes), same as the first treble part but consecutive equal tones are tied. This part is repeated. Playing time: 4' 47".

27 Pascal’s Triangle – Cool Stuff numbers divisible by 5 similar patterns exist for other numbers http://www.shodor. org/interactivate/ac tivities/pascal1/ http://www.shodor. org/interactivate/ac tivities/pascal1/

28 Pascal’s Triangle can also be seen in terms of combinations n = 0 n = 1 n = 2 n = 3 n = 4 n = 5 n = 6

29 Pascal’s Triangle - Summary symmetrical down the middle outside number is always 1 second diagonal values match the row numbers sum of each row is a power of 2  sum of nth row is 2 n  Begin count at 0 number inside a row is the sum of the two numbers above it

30 And one more thing… remember that for the inner numbers in the triangle, any number is the sum of the two numbers above it for example 4 + 6 = 10 this suggests the following: which is an example of Pascal’s Identity

31 For Example…

32 How can this help us solve our original problem? 1 5 15 35 70126 1 4 10 20 35 56 1 3 6 10 15 21 1 2 3 4 5 6 1 1 1 11 so by overlaying Pascal’s Triangle over the grid we can see that there are 126 ways to move from one corner to another

33 How many routes pass through the green square? to get to the green square, there are C(4,2) ways (6 ways) to get to the end from the green square there are C(5,3) ways (10 ways) in total there are 60 ways

34 How many routes do not pass through the green square? there are 60 ways that pass through the green square there are C(9,5) or 126 ways in total then there must be 126 – 60 = 66 paths that do not pass through the green square

35 MSIP / Home Learning Read the examples on pp. 281-287 The example starting on the bottom of page 287 is important page 289 #1, 2aceg, 6-8, 11-12


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