Probability Distributions and Expected Value Chapter 5.1 – Probability Distributions and Predictions Mathematics of Data Management (Nelson) MDM 4U

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 produce only 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}

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? 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 Outcome 3 4 5 6 7 8 9 … # ways 1

Probability Distribution for 3 Dice Outcome 3 4 5 6 7 8 9 10 # cases 1 15 21 28

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/are the most common outcome(s)? 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 1a: tossing 3 coins 0 heads 1 head 2 heads 3 heads P(X) ⅛ ⅜ 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

Example 1b: tossing 3 coins 0 heads 1 head 2 heads 3 heads P(X) ⅛ ⅜ 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

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 have 3 women

Example 2a cont’d: selecting a committee 0 women 1 woman 2 women 3 women P(X) 4/35 18/35 12/35 1/35 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

Example 2b: selecting a committee 0 women 1 woman 2 women 3 women P(X) 4/35 18/35 12/35 1/35 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)

MSIP / Homework p. 277 #1, 2, 3, 4, 5, 9, 12, 13