PROBABILITY

Objectives:

• Determine the sample space of an experiment.
• Find the probability of an event.
• Find the probability of mutual exclusive events.
• Find the probability of the complement of an event.
THE PROBABILITY OF AN EVENT

Probability is all about chance - the chance or likelihood that something has of happening. When we talk about probability we are talking about how likely something is to happen. For example, if we toss a fair coin, we say that the probability that it will land face up is 1/2.
Any happening whose result is uncertain is called an experiment. The possible results of an experiment are outcomes, the set of all possible outcomes of an experiment is the sample space of the experiment, and any subcollection of a sample space is an event

Example 1
a. When a six-sided die is tossed, the sample space can be represented by the numbers from 1 through 6, each having equally likely (or equally possible) outcomes:
S={1, 2, 3, 4, 5, 6}.

b. If you were to roll two dice and look at the sum of the two dice, then S = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }.

However, not all sample spaces are created equally. In fact, that last example is not. There is only one way that a sum of 2 can be rolled, a 1 on the first die and a 1 on the second die. There are four ways that a sum of five can be rolled: 1-4, 2-3, 3-2, 4-1 (don't be confused here, 1-4 is a 1 on the first die and a 4 on the second and is different than a 4 on the first and a 1 on the second. If it helps, pretend that you're rolling one die and a friend is rolling the other).

We want our sample spaces to be equally likely if at all possible.
To describe sample spaces in such a way that each outcome is equally likely you must sometimes distinguish between various outcomes in ways that appear artificial.

Example 2. Find the sample space for the following:
a. One coin is tossed.
b. Two coins are tossed.
c. Three coins are tossed.

Solution:
a. Because the coin will land either heads up (denoted by H) or tails up (denoted by T), the sample space is S={H, T}.

b. Because either coin can land heads up or tails up, the possible outcomes are
HH = heads oup on both coins
HT = heads up on first coin and tails up on second coin
TH = tails up on first coin and heads up on second coin
TT = tails up on both coins.

The sample space is S={HH, HT, TH, TT}. This list distinguishes between the two classes HT and TH, even though these two outcomes appear to be similar.

c. Following the notation of part (b), the sample space is
S={HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.

To calculate the probability of an event. count the numbers of outcomes in the event and in the sample space.  The number of outcomes in the event E is denoted by n(E), and the number of outcomes in the sample space S is denoted by n(S). The probability that the event E will occur is given by n(E)/n(S).

 The Probability of an Event If an event E has n(E) equally likely outcomes and its sample space has n(S) equally likely outcomes, the probability of event E is

Example 3 The probability of rolling a six on a single roll of a die is 1/6 because there is only 1 way to roll a six out of 6 ways it could be rolled.

Because the number of outcomes in an event must be less than or equal to the number of outcomes in the sample space (this is because the event is a subcollection of the sample space), the probability of an event must be a number between 0 and 1. That is, for any event E, it must be true that 0<= P(E) <= 1.

 Properties of the Probability of an Event Let E be an event that is a subset of a finite sample space S. 1. 0 <=P(E) <= 1 2. If P(E)=0, E cannot occur, and is called the impossible event. 3. If P(E) =1, E must occur and is called a certain event.

Example 4  Find the probability of getting a sum of 5 when rolling two dice.

Solution: The probability of getting a sum of 5 when rolling two dice is 4/36 = 1/9 because there are 4 ways to get a five and there are 36 ways to roll the dice (Fundamental Counting Principle - 6 ways to roll the first times 6 ways to roll the second).

Note
Do not make the mistake of saying that the probability of rolling a sum of 5 is 1/11 because there is one 5 out of a sample space of 11 sums (2 through 12). When the sample spaces are not equally likely, do not divide by the number in the sample space.

Example 5 Two coins are tossed. What is the probability that both land heads up?

Solution: Following the procedure in Example 2 (b), let E={HH} and S={HH, HT, TH, TT}. So n(E)=1 and n(S)=4. Then the probability of getting two heads is

P(E)=1/4

Example 6  A card is drawn from a standard deck of playing cards. What is the probability that it is an ace?

Solution: Because there are 52 cards in a standard deck and there are four aces (one in each suit), then n(E)=4 and n(S)=52,. so the probability of drawing an ace is

P(E)=4/52=1/13.

MUTUALLY EXCLUSIVE EVENTS

Two events A and B (from the same sample space S) are mutually exclusive if A and B have no outcomes in common. In the terminology of sets, the intersection  (or the product) of A and B is the empty set and
P(AB)=0
To find the probability that one or the other of two mutually exclusive events will occur, you can add their individual probabilities.

 Probability of the Union of Two Events If A and B are events in the same sample space S, the probability of A or B occurring is given by P(A or B)=P(A) + P(B) - P(AB) If A and B are mutually exclusive, then  P(A or B)=P(A) + P(B) This is called the addition rule.

Example 7     If two dice are tossed, the event A of rolling a total of 6 and the event B of rolling a total of 9 are mutually exclusive. Therefore, the probability of A or B occurring is given by

P(A or B)=P(A) + P(B)=5/36 + 4/36=9/36=1/4.

INDEPENDENT EVENTS

Two events are independent if the occurrence of one has no effect on the occurrence of the other. To find the probability that two independent events will occur, you can multiply their individual probabilities.

 Probability of Independent Events If A and B are independent events in the same sample space S, the probability that both A and B will occur is given by P(A and B)=P(A) *P(B) This is called the multiplication rule.

Example 8     A random number generator on a computer selects three integers from 1 to 20. What is the probability that all three numbers are less than or equal to 5?

Solution: The probability of selecting from all those 20 numbers a number between 1 and 5 is

P(A)=5/20= 1/4.

So, the probability that all three numbers are less than or equal to 5 is

P(A)*P(A)*P(A)= 1/64.

Example 9   A coin is thrown twice. What is the probability of throwing 2 heads?

Solution: You need first to work out how many possible outcomes there are and if these possible outcomes are equally probable.
Since there are two sides to a coin, the probability of throwing a head is 1 out of 2 (or 1/2 or 0.5).  The probability of throwing a tail is also 1/2.
No toss of the coin affects the next so the probabilities for each throw are equal.

To calculate the probability of one outcome followed by another, use multiplication.

Probability of throwing a head followed by a head will be
1/2 * 1/2=1/4

THE COMPLEMENT OF AN EVENT

The complement of an event A is the collection of all outcomes in the sample space that are not in A. The complement of A is denoted by A'. Because P(A or A')=1 and because A and A' are mutually exclusive, it follows that
P(A')=1-P(A).

For instance, if the probability of winning a certain game is

P(A)=1/4

the probability of losing the game is P(A')=1-1/4=3/4.

 Probability of a Complement Ley A be an event in the sample space S and let A' be its complement. If the probability of A is P(A), then the probability of its complement is P(A')=1-P(A) This is called the multiplication rule.

Example 10  If two dice are rolled, what is the probability that the sum is not 7?

Solution: Because "a sum of 7" and "a sum that is not 7" are mutually exclusive events, we can find the required probability by computing first the probability of the complementary event, that is P(sum is 7).
Let E represent the event of rolling a sum of 7. Then,

E={(1,6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) }

Thus P(E)=6/36=1/6. Therefore,

P(sum is not 7)=1-P(E)=1-1/6=5/6.

Example 11  Suppose you draw one card from a standard deck. What is the probability of obtaining a number card or a heart?

Solution: The sample space consists of 52 possible outcomes. Let H be the event "heart drawn" and N be the event "number card" drawn.
A standard deck contains 13 hearts and 36 number cards, but 9 of the number cards are also hearts.. So

P(H or N)=P(H)+P(N)-P(H and N)=13/52+36/52-9/52=10/13.

DEPENDENT EVENTS

Two events are dependent if the occurrence of one of the events affects the probability of the occurrence of the other.
Consider a box that contains seven white, five green, and four blue pieces of chalk. If two pieces of chalk are drawn from the box, and the first piece is not replaced before the second piece is drawn, the outcome of the first selection affects the outcome of the second drawing.The probability of drawing a piece of blue chalk the first time is 4/16 or 1/4. If a piece of blue chalk is drawn first, then three of the remaining pieces of chalk are blue, and the probability of getting a piece of blue chalk on the second draw is 3/15 or 1/5.

The probability that two dependent events will happen may be expressed as follows:

If pl is the probability that event A will happen and, after A has happened, p2 is the probability that event B will happen, then the probability that the events will occur in the order A,B is p1*p2, that is, the product of their respective probabilities.

CONDITIONAL PROBABILITY

The probability that one event will occur, given that some event has occured or is certain to occur, is a condtional probability.

The conditional probability of A given that B has occured is

P(A|B)=P(AB) / P(B).

Note: The notation P(A|B) is read "the probability of A given B".

If A and B are two independent events, then P(A|B)=P(A) and P(B|A)=P(B).

Example 12 Suppose a political survey has been done and information on two variables has been recorded: the age of the voters and whether they favor or don't favor a candidate. The results of the survey are given in the following table:

 Age favor Not Favor Undecided Total 18-35 213 197 103 513 35-50 193 184 67 444 Over 50 144 219 83 446 Total 550 600 253 1403

If one of the voters is selected at random:
a) What is the probability that they will favor a candidate?

Let A = favor the candidate. Then
P(A)= (number of persons who favor a candidate)/ (total number of persons)
P(A)= (213 + 193 + 144) / 1403
P(A)= 550 / 1403 = .3920

b) What is the probability that they will be between 35 and 50 years old?

Let B= age between 35 and 50. Then
P(B)= (number of persons whose age is between 35 and 50 years)/ total numbers of persons in the survey
P(B)= (193 + 184 67) / 1403
P(B)= 444/ 1403= .316

c) What is the probability that they will be over 50 years old and they will favor a candidate?

Let A = age over 50   and B= favor the candidate. Obviously A and B are not independent events.

P(age over 50 and favor a candidate)= P(AB) = (number of persons that are over 50 and favor a candidate)/ total number of people
P(AB)= 144/1403 = .1026

d) What is the probability that they wil be over 50 years?

Let A= age over 50 and A' = age not over 50. Then

P(A')= 1- P(A)
P(A)= number of persons that are over 50 / total number of persons in the survey
P(A)= 446/ 1402 =.3179

Therefore, P(A')= 1- .3179 = .6821.

e) If the respondent is between the ages of 18 and 35, what is the probability that they will favor the candidate?

Let A=favor the candidate and B= age between 18 and 35.
P(favor the candidate | age between 18 and 35) = P(A|B)= P(AB) / P(B)..
But P(AB) = (number of persons satisfying the conditions A and B)/ total number of people in the survey.
= 213/ 1403 = .1518
P(B)= (total number of persons whose ages are between 18 and 35)/ (total number of people in the survey)
P(B)= (213 + 197 + 103)/ 1403 = 513/ 1403 = .3656
Then P(A|B)= 0.1518/ 0.3656 = 0.4152.

f) If the respondent does not favor the candidate, what is the probability that they will be between 35 and 50?

Let A =not favor the candidate and B= age between 35 and 50.
Then P(age between 35 and 50 | does not favor the candidate) = P(B|A) = P(BA) / P(A)..
P(BA)= (number  of persons satisfying the conditions A and B) / (total number of people in the survey)
= 184 // 1403 =0.1311
P(A) = (number of persons that does not favor the candidate)/ (total number of people in the survey)
= (197 + 184 +219)/ 1403 =0.4277
Then P(B|A)= 0.1311/ 0.4277= 0.3065.

EXERCISES

1. Two balls are to be drawn successively from a bag known to contain only yellow balls and purple balls. List a sample space for the experiment.

2. Each letter in the word flower is written on a card and the cards are shuffled. List a sample space for the outcome of drawing one card.

3. The faces of a cube are marked with the letters A, A, B, C, D, E. If the cube is tossed, what is the probability that an A will turn up?

4. A committee of three is to be chosen from ten girls. If Ann, Betty and Carol are among the group of ten girls, what is the probability that all three of them will be on the committee?

5. If six cards are drawn at random from a deck of 52 cards, what is the probability that they are all spades?

6..In Hillcross High School there are 300 freshmen, 280 sophomores, 275 juniors, and 256 seniors. What is the probability that a student selected at random will be (a) a freshman? (b) a sophomore? (c) a junior? (d) a senior?

7.   If a bag contains four blue marbles, six yellow marbles, and five green marbles, what is the probability that in one drawing a person will pick either a blue marble or a green marble?

8. 1.) Are the following pairs of events mutually exclusive?
____a.) Living in New Haven and working in New York.
____b.) Being a freshman and being a junior in high school.
____c.) Being a professor and being an author of a book.
____d.) Drawing a red card and drawing the ace of spades.
____e.) Drawing a face card and drawing the six of hearts from a normal deck of cards.

9. When a card is drawn at random from a normal deck of 52 cards, what is the probability that it will be either an ace or a spade?

10. A bag contains five green marbles, four yellow marbles, and nine white marbles. If two marbles are drawn in succession, and the first marble is not replaced before the second is drawn, what is the probability that:
____a.) the second marble is yellow, if the first marble drawn is green?
____b.) the second marble is white, if the first marble drawn is yellow?
____c.) both marbles are green?
____d.) both marbles are yellow?
____e.) both marbles are white?

11. If five coins are tossed, what is the probability that all five coins will turn up heads?

12.  Find the probability that a five-card poker hand will be a full house (three of a kind and two of a kind).