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Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin twice is an example of an experiment.
A result of an experiment is called an outcome. The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter S is used to denote the sample space. For example, if you flip one fair coin, S = { H , T } where H = heads and T = tails are the outcomes.
An event is any combination of outcomes. Upper case letters like A and B represent events. For example, if the experiment is to flip one fair coin, event A might be getting at most one head. The probability of an event A is written P ( A ).
The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, zero and one and all numbers between these values). P ( A ) = 0 means the event A can never happen. P ( A ) = 1 means the event A always happens. P ( A ) = 0.5 means the event A is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads).
Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head ( H ) and a Tail ( T ) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.
To calculate the probability of an event A when all outcomes in the sample space are equally likely , count the number of outcomes for event A and divide by the total number of outcomes in the sample space. For example, if you toss a fair dime and a fair nickel, the sample space is { HH , TH , HT , TT } where T = tails and H = heads. The sample space has four outcomes. A = getting one head. There are two outcomes that meet this condition { HT , TH }, so P ( A ) = 24 = 0.5.
Suppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event E = rolling a number that is at least five. There are two outcomes {5, 6}. P ( E ) = 26. If you were to roll the die only a few times, you would not be surprised if your observed results did not match the probability. If you were to roll the die a very large number of times, you would expect that, overall, 26 of the rolls would result in an outcome of "at least five". You would not expect exactly 26. The long-term relative frequency of obtaining this result would approach the theoretical probability of 26 as the number of repetitions grows larger and larger.
This important characteristic of probability experiments is known as the law of large numbers which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.)
It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair, or biased. Two math professors in Europe had their statistics students test the Belgian one Euro coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44% of the time. The data seem to show that the coin is not a fair coin; more repetitions would
be helpful to draw a more accurate conclusion about such bias. Some dice may be biased. Look at the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely.
"∪" Event: The Union An outcome is in the event A ∪ B if the outcome is in A or is in B or is in both A and B. For example, let A =
{1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}. A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are NOT listed twice.
a. S =
Let event A = the even numbers and event B = numbers greater than 13.
b. A = , B =
c. P ( A ) = , P ( B ) =
d. A ∩ B = , A OR B =
e. P ( A ∩ B ) = , P ( A ∪ B ) =
f. A′ = , P ( A′ ) =
g. P ( A ) + P ( A′ ) =
h. P ( A | B ) = , P ( B | A ) = ; are the probabilities equal?
Solution 3.
a. S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}
b. A = {2, 4, 6, 8, 10, 12, 14, 16, 18}, B = {14, 15, 16, 17, 18, 19}
c. P ( A ) = 919 , P ( B ) = 619
d. A ∩ B = {14,16,18}, A OR B = {2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19}
e. P ( A ∩ B ) = 319 , P ( A ∪ B ) = 1219
f. A′ = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19; P ( A′ ) = 1019
g. P ( A ) + P ( A′ ) = 1 (919 + 1019 = 1)
h. P ( A | B ) = P ( A ∩ B ) P ( B ) = 36, P ( B | A ) = P ( A ∩ B ) P ( A ) = 39, NoTry
It 3.
The sample space S is all the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)).
a. S =
Let event A = the sum is even and event B = the first number is prime.
b. A = , B =
c. P ( A ) = , P ( B ) =
d. A ∩ B = , A ∪ B =
e. P ( A ∩ B ) = , P ( A ∪ B ) =
f. B′ = , P ( B′ ) =
g. P ( A ) + P ( A′ ) =
h. P ( A | B ) = , P ( B | A ) = ; are the probabilities equal?
Example 3.
A fair, six-sided die is rolled. Describe the sample space S , identify each of the following events with a subset of S and compute its probability (an outcome is the number of dots that show up).
a. Event T = the outcome is two. b. Event A = the outcome is an even number. c. Event B = the outcome is less than four. d. The complement of A. e. A | B f. B | A g. A ∩ B h. A ∪ B i. A ∪ B′ j. Event N = the outcome is a prime number. k. Event I = the outcome is seven.
Solution 3.
Table 3.
Let’s denote the events M = the subject is male, F = the subject is female, R = the subject is right-handed, L = the subject is left- handed. Compute the following probabilities:
a. P ( M ) b. P ( F ) c. P ( R ) d. P ( L ) e. P ( M ∩ R ) f. P ( F ∩ L ) g. P ( M ∪ F ) h. P ( M ∪ R ) i. P ( F ∪ L ) j. P ( M' ) k. P ( R | M ) l. P ( F | L ) m. P ( L | F )
Solution 3.
a. P ( M ) = 0.
b. P ( F ) = 0.
c. P ( R ) = 0.
d. P ( L ) = 0.
e. P ( M ∩ R ) = 0.
f. P ( F ∩ L ) = 0.
g. P ( M ∪ F ) = 1
h. P ( M ∪ R ) = 0.
i. P ( F ∪ L ) = 0.
j. P ( M' ) = 0.
k. P ( R | M ) = 0.8269 (rounded to four decimal places) l. P ( F | L ) = 0.3077 (rounded to four decimal places)
m. P ( L | F ) = 0.
Independent and mutually exclusive do not mean the same thing.
Independent Events
Two events are independent if one of the following are true:
• P ( A | B )= P ( A )
• P ( B | A )= P ( B )
• P ( A ∩ B )= P ( A ) P ( B )
Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent.
Sampling may be done with replacement or without replacement.
- With replacement : If each member of a population is replaced after it is picked, then that member has the possibility ofbeing chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.
- Without replacement : When sampling is done without replacement, each member of a population may be chosen only once.In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.
If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise.
Example 3.
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit.
a. Sampling with replacement:
Suppose you pick three cards with replacement. The first card you pick out of the 52 cards is the Q of spades. You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the 52-card deck. This time, the card is the Q of spades again. Your picks are { Q of spades, ten of clubs, Q of spades}. You have picked the Q of spades twice. You pick each card from the 52-card deck.
Example 3.
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs.
a. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are QS , 1 D , 1 C , QD. b. Suppose you pick four cards and put each card back before you pick the next card. Your cards are KH , 7 D , 6 D , KH.
Which of a. or b. did you sample with replacement and which did you sample without replacement?
Solution 3.
a. Without replacement; b. With replacement
Try It 3.
You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible?
Answer the same question for sampling with replacement.
a. QS , 1 D , 1 C , QD b. KH , 7 D , 6 D , KH c. QS , 7 D , 6 D , KS
Mutually Exclusive Events
A and B are mutually exclusive events if they cannot occur at the same time. Said another way, If A occurred then B cannot occur and vise-a-versa. This means that A and B do not share any outcomes and P ( A ∩ B )=0.
For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}. A ∩ B = {4, 5}. P ( A ∩ B )=210 and is not equal to zero. Therefore, A and B are not mutually exclusive. A and C do not have any numbers in common so P ( A ∩ C )=0. Therefore, A and C are mutually exclusive.
If it is not known whether A and B are mutually exclusive, assume they are not until you can show otherwise. The following examples illustrate these definitions and terms.
Example 3.
Flip two fair coins. (This is an experiment.)
The sample space is { HH , HT , TH , TT } where T = tails and H = heads. The outcomes are HH , HT , TH , and TT. The outcomes HT and TH are different. The HT means that the first coin showed heads and the second coin showed tails. The TH means that the first coin showed tails and the second coin showed heads.
- Let A = the event of getting at most one tail. (At most one tail means zero or one tail.) Then A can be written as { HH , HT , TH }. The outcome HH shows zero tails. HT and TH each show one tail.
- Let B = the event of getting all tails. B can be written as { TT }. B is the complement of A , so B = A′. Also, P ( A ) + P ( B ) = P ( A )
- The probabilities for A and for B are P ( A ) = 34 and P ( B ) = 14.
- Let C = the event of getting all heads. C = { HH }. Since B = { TT }, P ( B ∩ C )=0. B and C are mutually exclusive. ( B and C haveno members in common because you cannot have all tails and all heads at the same time.)
- Let D = event of getting more than one tail. D = { TT }. P ( D ) = 14
- Let E = event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) E = { HT , HH }. P ( E ) = 24
- Find the probability of getting at least one (one or two) tail in two flips. Let F = event of getting at least one tail in two flips.
F = { HT , TH , TT }. P ( F ) = 34
Example 3.
Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event A = a face is odd. Then A = {1, 3, 5}. Let event B = a face is even. Then B = {2, 4, 6}.
- Find the complement of A , A′. The complement of A , A′ , is B because A and B together make up the sample space. P ( A ) + P ( B ) = P ( A ) + P ( A′ ) = 1. Also, P ( A ) = 36 and P ( B ) = 36.
- Let event C = odd faces larger than two. Then C = {3, 5}. Let event D = all even faces smaller than five. Then D = {2, 4}. P ( C ∩ D )=0 because you cannot have an odd and even face at the same time. Therefore, C and D are mutually exclusive events.
- Let event E = all faces less than five. E = {1, 2, 3, 4}.
Are C and E mutually exclusive events? (Answer yes or no.) Why or why not?
Solution 3.
No. C = {3, 5} and E = {1, 2, 3, 4}. P ( C ∩ E )=16. To be mutually exclusive, P ( C ∩ E ) must be zero.
- Find P ( C | A ). This is a conditional probability. Recall that the event C is {3, 5} and event A is {1, 3, 5}. To find P ( C | A ), find the probability of C using the sample space A. You have reduced the sample space from the original sample space {1, 2, 3, 4,5, 6} to {1, 3, 5}. So, P ( C ∣∣ A )=23.
Try It 3.
Let event A = learning Spanish. Let event B = learning German. Then A ∩ B = learning Spanish and German. Suppose P ( A )=0.4 and
P ( B )=0.2. P ( A ∩ B )=0.08. Are events A and B independent? Hint: You must show ONE of the following:
• P ( A | B )= P ( A )
• P ( B | A )= P ( B )
• P ( A ∩ B )= P ( A ) P ( B )
Example 3.
Let event G = taking a math class. Let event H = taking a science class. Then, G ∩ H = taking a math class and a science class. Suppose P ( G )=0.6, P ( H )=0.5, and P ( G ∩ H )=0.3. Are G and H independent?
If G and H are independent, then you must show ONE of the following:
• P ( G | H )= P ( G )
• P ( H | G )= P ( H )
• P ( G ∩ H )= P ( G ) P ( H )
NOTE
The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information.
a. Show that P ( G | H )= P ( G ).
Solution 3.
P ( G ∣∣∣ H )= P ( G ∩ H ) P ( H )=0.30.5=0.6= P ( G )
b. Show P ( G ∩ H )= P ( G ) P ( H ).
Solution 3.
Suppose P ( C )=0.75, P ( D )=0.3, P ( C | D )=0.75 and P ( C ∩ D )=0.225.
Justify your answers to the following questions numerically.
a. Are C and D independent? b. Are C and D mutually exclusive? c. What is P ( D | C )?
Solution 3.
a. Yes, because P ( C | D )= P ( C ). b. No, because P ( C ∩ D ) is not equal to zero.
c. P ( D ∣∣∣ C )= P ( C ∩ D ) P ( C )=0.2250.75=0.
Try It 3.
A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that
P ( B )=0.40, P ( D )=0.30 and P ( B ∩ D )=0.20.
a. Find P ( B | D ). b. Find P ( D | B ). c. Are B and D independent? d. Are B and D mutually exclusive?
Example 3.
In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.
Let R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn.
The sample space S = R 1, R 2, R 3, B 1, B 2, B 3, B 4, B 5. S has eight outcomes.
- P ( R )=38. P ( B )=58. P ( R ∩ B )=0. (You cannot draw one card that is both red and blue.)
- P ( E )=38. (There are three even-numbered cards, R 2, B 2, and B 4.)
- P ( E ∣∣ B )=25. (There are five blue cards: B 1, B 2, B 3, B 4, and B 5. Out of the blue cards, there are two even cards; B 2 and B 4.)
- P ( B ∣∣ E )=23. (There are three even-numbered cards: R 2, B 2, and B 4. Out of the even-numbered cards, to are blue; B 2 and B 4.)
- The events R and B are mutually exclusive because P ( R ∩ B )=0.
- Let G = card with a number greater than 3. G = { B 4, B 5}. P ( G )=28. Let H = blue card numbered between one and four, inclusive. H = { B 1, B 2, B 3, B 4}. P ( G ∣∣ H )=14. (The only card in H that has a number greater than three is B 4.) Since 28 = 14, P ( G )= P ( G | H ), which means that G and H are independent.
Try It 3.
In a basketball arena,
- 70% of the fans are rooting for the home team.
- 25% of the fans are wearing blue.
- 20% of the fans are wearing blue and are rooting for the away team.
- Of the fans rooting for the away team, 67% are wearing blue.
Let A be the event that a fan is rooting for the away team. Let B be the event that a fan is wearing blue.
Are the events of rooting for the away team and wearing blue independent? Are they mutually exclusive?
Example 3.
outcomes.
b. Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5 or 6 dots on a side). The outcomes are. Count theoutcomes. There are outcomes. c. Multiply the two numbers of outcomes. The answer is. d. If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer to c is the number of outcomes (sizeof the sample space). What are the outcomes? (Hint: Two of the outcomes are H 1 and T 6.) e. Event A = heads ( H ) on the coin followed by an even number (2, 4, 6) on the die.
A = { }. Find P ( A ).
f. Event B = heads on the coin followed by a three on the die. B = { }. Find P ( B ). g. Are A and B mutually exclusive? (Hint: What is P ( A ∩ B )? If P ( A ∩ B )=0, then A and B are mutually exclusive.) h. Are A and B independent? (Hint: Is P ( A ∩ B )= P ( A ) P ( B )? If P ( A ∩ B )= P ( A ) P ( B ), then A and B are independent. If not, then they are dependent).
Solution 3.
a. H and T ; 2
b. 1, 2, 3, 4, 5, 6; 6
c. 2(6) = 12
d. T 1, T 2, T 3, T 4, T 5, T 6, H 1, H 2, H 3, H 4, H 5, H 6
e. A = { H 2, H 4, H 6}; P ( A ) = 312
f. B = { H 3}; P ( B ) = 112
g. Yes, because P ( A ∩ B )= h. P ( A ∩ B )=0. P ( A ) P ( B )=(312). P ( A ∩ B )does not equal P ( A ) P ( B ),so A and B are dependent.
Try It 3.
A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let T be the event of getting the white ball twice, F the event of picking the white ball first, S the event of picking the white ball in the second drawing.
a. Compute P ( T ). b. Compute P ( T | F ). c. Are T and F independent?. d. Are F and S mutually exclusive? e. Are F and S independent?
