Probability Distributions and Expected Value Chapter 5.1 – Probability Distributions and Predictions Mathematics of Data Management (Nelson) MDM 4U Authors: Gary Greer (with K. Myers)

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

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

Probability Distribution for 2 Dice

What would a probability distribution graph for three dice look like? lets try it! Using three dice, figure out how many possible cases there are now find out how many possible ways there are to create each of the possible cases fill in a table like the one below now you can make your graph Outcome … # cases1

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

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

Example: 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

Example: 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

Selecting a Team 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) have no women C(4,2) x C(3,1) have one woman C(4,1) x C(3,2) have 2 women C(3,3) have 3 women

Example: 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 + 1/35 = 31/35 is there an easier way???? X0 women1 woman2 women3 women P(X)4/3518/3512/351/35

Example: 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

Exercises / Homework Homework: page 277 #1, 2, 3, 4, 5, 9, 12, 13

Pascal’s Triangle and the Binomial Theorem Chapter 5.2 – Probability Distributions and Predictions Mathematics of Data Management (Nelson) MDM 4U Authors: Gary Greer (with K. Myers)

How many routes are there to the top right-hand corner? you need to move up 4 spaces and over 5 spaces the total routes can be calculated with C(9,5) or C(9,4) = 126 ways

The Binomial Theorem the term (a + b) can be expanded:  (a + b) 0 = 1  (a + b) 1 = a + b  (a + b) 2 = a 2 + 2ab + b 2  (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3  (a + b) 4 = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4  Blaise Pascal (for whom the Pascal computer language is named) noted that there are patterns of expansion, and from this he developed what we now know as Pascal’s Triangle. He also invented the second mechanical calculator

Pascal’s Triangle the outer values are always 1 the inner values are determined by adding the values of the two values diagonally above

Pascal’s Triangle sum of each row is 1 = = = = = = = 2 6

Pascal’s Triangle Uses? binomial theorem combinations!  choose 2 items from 5  go to the 5 th row, the 2 nd number = 10 (always start counting at 0)  so it can be used to find combinations modeling the electrons in each shell of an atom

Pascal’s Triangle – Cool Stuff 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 enna/9349/pascal.mid

Pascal’s Triangle – Cool Stuff numbers divisible by 5 similar patterns exist for other numbers org/interactivate/ac tivities/pascal1/

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

Pascal’s Triangle symmetrical down the middle outside 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 number inside a row is the sum of the two numbers above it

So what does this have to do with the Binomial Theorem remember that:  (a + b) 4 = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4  and the triangle’s 4 th row is so Pascal’s Triangle allows you to predict the coefficients in the binomial expansion notice also that the exponents on the variables also form a predictable pattern with the exponents of each term having a sum of n

The Binomial Theorem

A Binomial Expansion lets expand (x + y) 4

Another Binomial Expansion lets expand (a + 4) 5

Some Binomial Examples what is the 6 th term (a + b) 9 ? don’t forget that when you find the 6 th term, r = 5 what is the 11 th term of (2x + 4) 12

Finding the term…

Look at the triangle in a different way r0 r1 r2 r3 r4 r5 n = 0 1 n = n = n = n = n = n = for a binomial expansion of (a + b) 5, the term for r = 3 has a coefficient of 10

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 = 10 this suggests the following which provides an example of Pascal’s Identity

For Example…

How can this help us solve our original problem? so by overlaying Pascal’s Triangle over the grid we can see that there are 126 ways to move from one corner to another

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

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 paths that do not pass through the green square

Exercises / Homework Homework: read the examples on pages , in particular the example starting on the bottom of page 287 is important page 289 #1, 2 a c e g, 3, 4, 5, 6, 8, 9, 11, 13