stat 230 midterm 1 exam-aid october 11, 2011

STAT 230 MIDTERM 1 EXAM-AIDOctober 11, 2011

Introduction• Steve Hobbs• 3A Actuarial Science and Statistics Double Major• Second Time Volunteering for SOS• Hardcore Gamer• Hobbs_Steve@hotmail.com

Outline1. Sample Spaces and Probability2. Counting Arguments3. Unions and Intersections of Events4. Independence and Mutual Exclusivity5. Conditional Probability

Sample Space

• Sample Space (S): A set of distinct outcomes for an experiment or process, such that exactly one outcome occurs on each trial – e.g. the sample space for the outcomes on one roll

of a die = {1, 2, 3, 4, 5, 6}

• Usually the denominator for a probability

Simple Event• Simple Event - contains only one point

– e.g. rolling a 1 is a simple event within our die sample space, since the event contains only the point {1}

• Probability of a simple event – if {a1, a2, …, ak} make up a sample space, then the probability of ai occurring for i = 1, 2, …, k follows– 0 ≤ P(ai) ≤ 1

– Sum of all P(ai’s) =1

– If all events are equally likely, then P(ai) = 1/k

Compound Event• Compound Event – contains two or more simple

events– e.g. rolling a 1 or a 2 is a compound event within our die

sample space, since the event contains the points {1, 2}

• Probability of compound event - sum of the probabilities for all the simple events that make up the compound event– For an event A, P(A) = number of outcomes in A

total number of outcomes in S– P(S) = 1

Addition Rule

• The Addition Rule: If we can do A in p ways and B in q ways, then we can do either A or B but not both in p + q ways– For events with no overlap (mutually exclusive)– e.g. if we can roll a 1 in one way and we can roll a

2 or 3 in two ways, then we can roll a 1, 2, or 3 in three ways

Multiplication Rule

• The Multiplication Rule: If we can do job A in p ways and for each of these ways we can do job B in q ways, then we can do both A and B in p x q ways– e.g. if we roll an even number on our first roll of

the die (3 ways) and an odd number on our second roll of the die (3 ways), the total number of {even, odd} pairs = 3 x 3 = 9

Problem 1Rebecca Black and her two friends each randomly choose a car for their ride home. There are four cars from which to choose, and a capacity of six people per car. Three of the cars are white, and one is black.

a) What is the probability that they all end up in the black car?b) What is the probability that they all end up in the same car?c) What is the probability that they all end up in different cars?d) What is the probability that no one ends up in the black car?

Permutations

• The number of ways to arrange n distinct objects in a row is:

n! = n(n-1)(n-2)…(1)

– Note that we are arranging all n objects, not a subset of the n objects

Permutations

• The number of ways to arrange r objects selected from n distinct objects is:

n(r) = n! / (n-r)! = n(n-1)(n-2)…(n-r+1)

– For permutations, the order of arrangement matters!

Permutations

• The number of distinct arrangements of n objects when n1 are alike of one type, n2 alike of a second type, …, nk alike of a kth type (where n1 + n2 + … + nk = n) is:

n! (n1! x n2! x … x nk!)

Combinations

• The number of ways to choose r objects from n distinct objects is denoted by:

n n! r r! x (n-r)!

– The order of selection does not matter!– # of combinations ≤ # of permutations

=

Problem 2

Is the following a permutation or a combination? Write out the expression for the number of ways to:

a) Choose a President, VP, and Secretary from a group of 30 peopleb) Order a 3-topping pizza from a

selection of 11 toppings

Problem 3

The letters of the word EXCELLENT are arranged in a random order. Find the probability that:

a) The letter X occurs firstb) the same letter occurs at each endc) X, C, and N occur together, in any order

How many three letter combinations can be chosen containing at most one of each letter?

Replacement• With replacement: each time an object is selected

from a group of n objects, it is returned to the group of n objects– There are always n objects– Probabilities remain constant!

• Without replacement: each time an object is selected from the group of objects, it is removed from the group– The number of objects decreases after each trial– Probabilities change as well!

Problem 4

Three numbers are chosen from 1, …, 10. Find the probability that the numbers are drawn in increasing order if:

a) draws are made without replacementb) draws are made with replacement

Unions and Intersections

• If A B, then P(A) ≤ P(B)

• P(A) = 1 – P( A )

• P(A U B) = P (A) + P(B) – P(A B)– Union of A and B contains events in A or B (or both)– Intersection of A and B contains only events in both

A and B

De Morgan’s Laws

1. A U B = A B

2. A B = A U B

Problem 5

According to a survey of people in Stat 230, 55% are female, 55% are intelligent, and 15% are male and not intelligent. What percent are female and intelligent?

Independence and Mutual Exclusivity

• A and B are independent if and only if:

P(A B) = P(A) x P(B)

• A and B are mutually exclusive (disjoint) if and only if:

P(A B) = 0

How can events be both independent and mutually exclusive?

Problem 6

Given P(A) = 0.3, P(B) = 0.35, calculate P(A U B) given:

i) A and B are independentii) A and B are mutually exclusiveiii) A is a subset of B

Conditional Probability

• The conditional probability of event B, given that event A occurs, is:

P(B|A) = P (B A) P(A)

– If A and B are independent: P(B | A) = P(B)– P(B|A) = 1 – P(B|A)

• Multiplication Rule:P (B A) = P(A) x P(B|A) (from above)P(C B A) = P(A) x P(B|A) x P(C|A B)

Problem 7

A fair coin is tossed 4 times. Find the probability that, if at least 1 Tail occurs, exactly 1 Tail occurs.

Problem 8

Gilbert Gambler chooses two cards from a standard deck without replacement. What is the probability that he chooses a Jack and then a Queen, given that the first card chosen was a face card (Jack, Queen, King)?

Let A be the event that the first card is a face card Let B be the event that he chooses a Jack then a Queen

Bayes’ Theorem

• Bayes’ Theorem:P(A | B) = P(B | A) * P(A)

P(B)

– Combines multiplication rule and conditional probability

Problem 9

80% of Laurier students like chocolate. 75% of Laurier students need help with their math homework. Of those Laurier students who like chocolate, only 10% do not need help with their math homework. What percentage of Laurier students who need help with their math homework like chocolate?

Partition Rule

• If A1, A2, …, Ak is a partition of the sample space S into mutually exclusive events, then for an event B in S:P(B) = P(B A1) + P(B A2) + … + P(B Ak)

= P(B |Ai) x P(Ai) (multiplication rule)

– Simplest form: P(B) = P(B A) + P(B A )

Problem 10

40% of Waterloo students are female. The probability that a female student is “cool” is 75%, whereas the probability that a male student is “cool” is 30%. What is the probability that a randomly selected cool student is female?