Review - Week 6. It is often helpful to use tree diagrams to visualize conditional events and probabilities.

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Warren Watts
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1 Review - Week 6 Read: Chapters Review: More Probability Rules For any two events A and B, A = A) + - A If A and B are disjoint then A =0 If P ( > 0, then A A = or alternatively P ( A = A Two events A and B are independent when A =A) It is often helpful to use tree diagrams to visualize conditional events and probabilities Exercise 1: (155) An inspection of dorm rooms on campus showed that 30% had refrigerators, 60% had TVs, and 18% had both a TV and a refrigerator What s the probability that a randomly selected dorm room has (a) a TV but no refrigerator? (b) a TV or a refrigerator, but not both? (c) neither a TV nor a refrigerator? Exercise : (157) You draw a card at random from a standard deck of 5 cards Find each of the following conditional probabilities: (a) The card is a heart, given that it is red (b) The card is red, given it is a heart (c) The card is an ace, given that it is red (d) The card is a queen, given that it is a face card Exercise 3: (1513) You are dealt a hand of three cards, one at a time Find the probability of each of the following (a) The first heart you get is the third card dealt (b) Your cards are all red (that is, all diamonds or hearts) (c) You get no spades (d) You have at least one ace

2 Exercise 4: When a woman gives birth, she has twins with probability 1/80; otherwise assume she had a single birth (For simplicity, ignore complications such as fertility drugs, triplets, quadruplets, etc) A woman gives birth twice (a) What is the probability these births give rise to exactly three children? (b) What is the probability that she had twins the first time given that she had twins the second time? Are these two events independent? Review: Random Variables A random variable is a variable whose value is a numerical outcome of a random phenomenon X is a discrete random variable if it takes a finite number of possible values The probability model of X, lists these values and their probabilities Value of X x 1 x x k Probability p 1 p p k Every probability p i is a number between 0 and 1 p 1 + p + + p k =1 X is a continuous random variable if it takes all possible values in an interval of numbers The probability model of X is described by a smooth curve The probability model for a continuous random variable assigns probabilities to intervals of outcomes rather than to individual outcomes Each individual outcome is assigned probability 0 The Normal model can be used to describe a continuous random variable In this case we say that the random variable follows the normal model The expected value, or mean, of a random variable describes its theoretical long-run average value The mean of X is given by µ = E ( = x1 p1 + x p + + xk pk = xi pi The variance of X is σ = Var( = ( x µ ) p + ( x µ ) p + + ( xk µ ) pk = ( xi µ ) pi 1 The standard deviation of X, σ, is the square root of the variance 1 In order to calculate the mean and variance for continuous random variables we need calculus

3 Rules for means: If c is a constant, then E ( X + c) = E( + c If a is a constant, then E ( a = ae( If X and Y are random variables, then E ( X + = E( + E( and E( X = E( E( Two random variables X and Y are independent if every event associated with X alone is independent of any event associated with Y alone Rules for Variance: If c is a constant, then Var ( X + c) = Var( If a is a constant, then Var ( a = a Var( If X and Y are independent then Var ( X + = Var( + Var( and Var ( X = Var( + Var( Exercise 1: Suppose X follows a normal model with mean 10 and standard deviation 34 Find the following probabilities: (a) X 84) (b) X 108) (c) 77 X 13) Exercise : The number of students who attend office hours is a random variable X, with P ( X = 0) = 01, X = 1) = 0 6 and P ( X = ) = 0 3 Calculate the mean and variance of X Exercise 3: Suppose X is a random variable with mean and standard deviation 5 Further suppose that Y is a random variable with mean 4 and standard deviation 3 You may assume that X and Y are independent Find the mean and variance of the random variables: (a) X-5 (b) X (c) X-Y (d) (X+Y)/ Exercise 4: Suppose X 1 is a random variable with mean µ and standard deviation σ Further suppose that X is a random variable with the same mean and standard deviation You may X 1 + X assume that X 1 and X are independent Find the mean and standard deviation of

4 Examples of the rules for means and variances The following examples are meant to illustrate the rules for means and variances in practice Ex 1: Suppose you flip two coins Let X = number of heads The probability model for X is given by: Value of X 0 1 E ( = 05(0) + 05(1) + 05() = 1 Var ( = (0 1) 05 + (1 1) 05 + ( 1) 05 = 05 Flip two more coins Suppose you get 1 point for each head and you start with two points Let Y = number of points We can express this random variable as Y = X+ The probability model for Y is: Value of Y 3 4 E ( = 05() + 05(3) + 05(4) = 3 Var ( = ( 3) 05 + (3 3) 05 + (4 3) 05 = 05 Hence, E ( X + ) = E( + and Var ( X + ) = Var( Rule: If c is a constant, then E ( X + c) = E( + c and Var ( X + c) = Var( Ex : Flip a coin twice Suppose you get points for each head Let Y = number of points We can express this random variable as Y = X The probability model for Y is given by: Value of Y 0 4 E ( = and Var ( = Hence, E ( X ) = E( and Var ( = 4Var( Rule: If a is a constant, then E ( a = ae( and Var ( a = a Var(

5 Ex 3: Suppose persons A and B each flip a coin twice Let X = number of heads for person A Y = number of heads for person B Let Z = X+Y = total number of heads X can take values 0, 1 or Y can take values 0, 1 or Z can take values 0, 1,, 3 or 4 The probability model for Z is given by (verify this at home): Value of Z Probability 1/16 4/16 6/16 4/16 1/16 E ( Z) = and Var ( Z) = 1 Hence, E ( X + = E( + E( = = and Var ( X + = Var( + Var( = = 1 Rule: If X and Y are random variables, then E ( X + = E( + E( If they are independent then Var ( X + = Var( + Var( Let Z = X-Y = difference in number of heads X can take values 0, 1 or Y can take values 0, 1 or Z can take values -, -1, 0, 1 or The probability model for Z is given by (verify this at home): Value of Z Probability 1/16 4/16 6/16 4/16 1/16 E ( Z) = 0 and Var ( Z) = 1 Hence, E ( X = E( E( = 1 1 = 0 and Var ( X = Var( + Var( = = 1 Rule: If X and Y are random variables, then E( X = E( E( If they are independent then Var ( X = Var( + Var(

Name: Date: Use the following to answer questions 2-4:

Name: Date: Use the following to answer questions 2-4: Name: Date: 1. A phenomenon is observed many, many times under identical conditions. The proportion of times a particular event A occurs is recorded. What does this proportion represent? A) The probability