19 3.2 Independent and Mutually Exclusive Events
Independent and mutually exclusive do not mean the same thing.
Independent Events
Two events are independent if the following are true:
 P(AB) = P(A)
 P(BA) = P(B)
 P(A AND 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 of being 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.
You have a fair, wellshuffled 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.
 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 52card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the 52card 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 52card deck.  Sampling without replacement:Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is the
K of hearts. You put this card aside and pick the second card from the 51 cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining 50 cards in the deck. The third card is the J of spades. Your picks are {K of hearts, three of diamonds, J of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice.
Example 1
You have a fair, wellshuffled 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. Three cards are picked at random.
 Suppose you know that the picked cards are Q of spades, K of hearts and Q of spades. Can you decide if the sampling was with or without replacement?
Show Answer
Sampling with replacement
 Suppose you know that the picked cards are Q of spades, K of hearts, and J of spades. Can you decide if the sampling was with or without replacement?
Show Answer
No, we cannot tell if the sampling was with or without replacement.
Example 2
You have a fair, wellshuffled 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 you pick four cards, but do not put any cards back into the deck. Your cards are QS, 1D, 1C, QD.
 Suppose you pick four cards and put each card back before you pick the next card. Your cards are KH, 7D, 6D, KH.
Which of 1 or 2 did you sample with replacement and which did you sample without replacement?
Show Answer
 Without replacement
 With replacement
This video provides a brief lesson on finding the probability of independent events.
Try It
You have a fair, wellshuffled 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.
 QS, 7D, 6D, KS
 KH, 7D, 6D, KH
 QS, 1D, 1C, QD
Which of the following outcomes are possible for sampling without replacement?
Show Answer
Without replacement:
 QS, 7D, 6D, KS: Possible
 KH, 7D, 6D, KH: Impossible
 QS, 1D, 1C, QD: Possible
Which of the following outcomes are possible for sampling with replacement?
Show Answer
With replacement:
 QS, 7D, 6D, KS: Possible
 KH, 7D, 6D, KH: Possible
 QS, 1D, 1C, QD: Possible
Mutually Exclusive Events
A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A AND 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 AND B = {4, 5}. [latex]\displaystyle{P}{({A} \text{ AND } {B})}=\frac{{2}}{{10}}[/latex]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 AND 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 possible 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′.
P(A) + P(B) = P(A) + P(A′) = 1.  The probabilities for A = P(event A) = [latex]\frac{{3}}{{4}}[/latex].
 Let C = the event of getting all heads. C = {HH}. Are eevnt C and event B mutually exclusive?
Show Answer
Since B = {TT}, P(B AND C) = 0.
B and C are mutually exclusive. (B and C have no 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}. Find P(D).
Show Answer
P(D) = [latex]\frac{{1}}{{4}}[/latex]
 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.) Find P(E).
Show Answer
E = {HT,HH}.
P(E) = [latex]\frac{{2}}{{4}}[/latex]  Find the probability of getting at least one (one or two) tail in two flips.
Show Answer
Let F = event of getting at least one tail in two flips.
F = {HT, TH, TT}.
P(F) = [latex]\frac{{3}}{{4}}[/latex]
Try It
Draw two cards from a standard 52card deck with replacement. Find the probability of getting at least one black card.
Show Answer
The sample space of drawing two cards with replacement from a standard 52card deck with respect to color is {BB, BR, RB, RR}.
Event A = Getting at least one black card = {BB, BR, RB}
[latex]\displaystyle{P}{({A})}=\frac{{3}}{{4}}={0.75}[/latex]
Example 4
Flip two fair coins. Find the probabilities of the events.
 Let F = the event of getting at most one tail (zero or one tail).
 Let G = the event of getting two faces that are the same.
 Let H = the event of getting a head on the first flip followed by a head or tail on the second flip.
 Are F and G mutually exclusive?
 Let J = the event of getting all tails. Are J and H mutually exclusive?
Show Answer
The sample space is {HH, HT, TH, TT} where T = tails and H = heads.
 Zero (0) or one (1) tails occur when the outcomes HH, TH, HT show up. P(F) = [latex]\frac{{3}}{{4}}[/latex]
 Two faces are the same if HH or TT show up. P(G) = [latex]\frac{{2}}{{4}}[/latex]
 A head on the first flip followed by a head or tail on the second flip occurs when HH or HT show up. P(H) = [latex]\frac{{2}}{{4}}[/latex]
 F and G share HH so P(F AND G) is not equal to zero (0). F and G are not mutually exclusive.
 Getting all tails occurs when tails shows up on both coins (TT). H‘s outcomes are HH and HT.
J and H have nothing in common so P(J AND H) = 0. J and H are mutually exclusive.
This video provides two more examples of finding the probability of events that are mutually exclusive.
Try It
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).
 If event F = the event of getting the white ball twice, find P(F).
Show Answer
P(F) = [latex]\displaystyle\frac{{1}}{{4}}[/latex]
 If event G = the event of getting two balls of different colors, find P(G).
Show Answer
P(G) = [latex]\displaystyle\frac{{1}}{{2}}[/latex]
 If event H = the event of getting white on the first pick, find P(G).
Show Answer
P(H) = [latex]\displaystyle\frac{{1}}{{2}}[/latex]
 Are F and G mutually exclusive?
Show Answer
Yes
 Are G and H mutually exclusive?
Show Answer
No
Example 5
Roll one fair, sixsided 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′.
Show Answer
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) = [latex]\displaystyle\frac{{3}}{{6}}[/latex].
 Let event C = odd faces larger than two. Let event D = all even faces smaller than five.
Are C and D mutually exclusive events? Why?Show Answer
C = {3, 5}.D = {2, 4}.
P(C AND 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.
Are C and E mutually exclusive events? Why?Show Answer
C = {3, 5} and E = {1, 2, 3, 4}.
C and E = {3}
P(C AND E) = [latex]\frac{{1}}{{6}}[/latex].
No, event C and event E are not mutually exclusive events.  Find P(CA).
Show Answer
This is a conditional probability.
Recall that the event C is {3, 5} and event A is {1, 3, 5}.
To find P(CA), 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(CA) = [latex]\displaystyle\frac{{2}}{{3}}[/latex].
Try It
Let event A = learning Spanish. Let event B = learning German. Then A AND B = learning Spanish and German.
Suppose P(A) = 0.4 and P(B) = 0.2. P(A AND B) = 0.08.
Are events A and B independent?
Hint: You must show ONE of the following:

Show Answer
P(AB)=[latex]\frac{{{P}{({A}\text{ AND }{B})}}}{{{P}{({B})}}}=\frac{{0.08}}{{0.2}}={0.4}={P}{({A})}[/latex]
The events are independent because P(AB) = P(A).
Example 6
Let event G = taking a math class. Let event H = taking a science class. Then, G AND H = taking a math class and a science class. Suppose P(G) = 0.6, P(H) = 0.5, and P(G AND H) = 0.3. Are G and H independent?
Hint:If G and H are independent, then you must show ONE of the following:


Method 1:
Show that P(GH) = P(G).
P(GH) = [latex]\frac{\text{P(G and H)}}{\text{P(H)}}[/latex] = [latex]\frac{0.3}{0.5}[/latex] = 0.6
P(G) = 0.6
Since P(GH) = P(G), G and H are independent. 
Method 2:
Show P(G AND H) = P(G)P(H).
P(G and H) = 0.3
P(G) * P(H) = 0.6 * 0.5 = 0.3
Since P(G AND H) = P(G)P(H), G and H are independent. 
Method 3:
Show that P(HG) = P(H).
P(HG) = [latex]\frac{\text{P(H and G)}}{P(G)}[/latex] = [latex]\frac{0.3}{0.6}[/latex] = 0.5
P(H) = 0.5
Since P(HG) = P(H), G and H are independent.
Since G and H are independent, knowing that a person is taking a science class does not change the chance that he or she is taking a math class. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance he or she is taking math.
Try It
In a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4.
 R = a red marble
 G = a green marble
 O = an oddnumbered marble
 The sample space is S = {R1, R2, R3, R4, R5, R6, G1, G2, G3, G4}.
S has ten outcomes. What is P(G AND O)?
Show Answer
Event G and O = {G1, G3}
P(G and O) = [latex]\displaystyle\frac{{2}}{{10}}[/latex] = 0.2
Example 7
Let event C = taking an English class. Let event D = taking a speech class.
Suppose P(C) = 0.75, P(D) = 0.3, P(CD) = 0.75 and P(C AND D) = 0.225.
Justify your answers to the following questions numerically.
 Are C and D independent?
Show Answer
Yes, because P(CD) = P(C).
 Are C and D mutually exclusive?
Show Answer
No, because P(C AND D) is not equal to zero.
 What is P(DC)?
Show Answer
[latex]\displaystyle{P}{({D}{mid}{C})}=\frac{{{P}{({C} \text{ AND }{D})}}}{{{P}{({C})}}}=\frac{{0.225}}{{0.75}}={0.3}[/latex]
Try It
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 AND D) = 0.20.
 Find P(BD).
 Find P(DB).
 Are B and D independent?
 Are B and D mutually exclusive?
Show Answer
 P(BD) = 0.6667
 P(DB) = 0.5
 No
 No
Example 8
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 wellshuffled. 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 = evennumbered card is drawn.
The sample space S = R1, R2, R3, B1, B2, B3, B4, B5. S has eight outcomes.
 Find P(R).
Show Answer
P(R) = [latex]\frac{{3}}{{8}}[/latex].
 Find P(R and B).
Show Answer
P(R AND B) = 0. (You cannot draw one card that is both red and blue.)
 Find P(E).
Show Answer
P(E) = [latex]\frac{{3}}{{8}}[/latex]. (There are three evennumbered cards, R2, B2, and B4.)
 Find P(EB) .
Show Answer
P(EB) = [latex]\frac{\text{P(E and B)}}{P(B)}[/latex] = [latex]\frac{{2}}{{5}}[/latex]
(There are five blue cards: B1, B2, B3, B4, and B5.
Out of the blue cards, there are two even cards; B2 and B4.)  Find P(BE).
Show Answer
P(BE) = [latex]\frac{\text{P(E and B)}}{P(E)}[/latex] = [latex]\frac{{2}}{{3}}[/latex].
There are three evennumbered cards: R2, B2, and B4.
Out of the evennumbered cards, to are blue; B2 and B4.  Are events R and mutually exclusive?
Show Answer
The events R and B are mutually exclusive because P(R AND B) = 0.
 Let G = card with a number greater than 3. Find P(G).
Show Answer
G = {B4, B5}. P(G) = [latex]\frac{{2}}{{8}}[/latex], P(G) = P(GH), which means that G and H are independent.
Try It
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, 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?
Show Answer
 P(BA) = 0.67, P(B) = 0.25
So P(B) does not equal P(BA) which means that B and A are not independent (wearing blue and rooting for the away team are not independent).  They are also not mutually exclusive, because P(B AND A) = 0.20 [latex]\ne[/latex] 0.
Example 9
In a particular college class, 60% of the students are female. Fifty percent of all students in the class have long hair. Fortyfive percent of the students are female and have long hair. Of the female students, 75% have long hair. Let F be the event that a student is female. Let L be the event that a student has long hair. One student is picked randomly. Are the events of being female and having long hair independent?
 The following probabilities are given in this example:
 P(F) = 0.60; P(L) = 0.50
 P(F AND L) = 0.45
 P(LF) = 0.75
Solution 1:
Check if P(F AND L) = P(F) * P(L).
We are given that P(F AND L) = 0.45, but P(F) * P(L) = (0.60)(0.50) = 0.30.
P(F AND L) [latex]\ne[/latex] P(F)P(L), the events of being female and having long hair are not independent.
Solution 2
Check whether P(LF) equals P(L).
We are given that P(LF) = 0.75, but P(L) = 0.50.
Since P(LF) [latex]\ne[/latex] P(L), the events of being female and having long hair are not independent.
Interpretation of ResultsThe events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair. 
Try It
Mark is deciding which route to take to work. His choices are I = the Interstate and F = Fifth Street.
 P(I) = 0.44 and P(F) = 0.55
 P(I AND F) = 0 because Mark will take only one route to work.
What is the probability of P(I OR F)?
Show Answer
P(I AND F) = 0,
P(I OR F) = P(I) + P(F) – P(I AND F) = 0.44 + 0.56 – 0 = 1
Example 10
 Toss one fair coin (the coin has two sides, H and T). The outcomes are ________. Count the outcomes. There are ____ outcomes.
Show Answer
H and T; 2
 Toss one fair, sixsided die (the die has 1, 2, 3, 4, 5 or 6 dots on a side). The outcomes are ________________. Count the outcomes. There are ___ outcomes.
Show Answer
1, 2, 3, 4, 5, 6 ; 6
 Multiply the two numbers of outcomes. The answer is _______.
Show Answer
2(6) = 12
 If you flip one fair coin and follow it with the toss of one fair, sixsided die, the answer in three is the number of outcomes (size of the sample space). What are the outcomes? (Hint: Two of the outcomes are H1 and T6.)
Show Answer
T1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6
 Event A = heads (H) on the coin followed by an even number (2, 4, 6) on the die.
A = {_________________}. Find P(A).Show Answer
A = {H2, H4, H6}; P(A) = [latex]\frac{{3}}{{12}}[/latex]
 Event B = heads on the coin followed by a three on the die. B = {________}. Find P(B).
Show Answer
B = {H3}; P(B) = [latex]\frac{{1}}{{12}}[/latex]
 Are A and B mutually exclusive?
(Hint: What is P(A AND B)? If P(A AND B) = 0, then A and B are mutually exclusive.)Show Answer
Yes, because P(A AND B) = 0
 Are A and B independent?
(Hint: Is P(A AND B) = P(A)P(B)? If P(A AND B) = P(A)P(B), then A and B are independent. If not, then they are dependent).Show Answer
P(A AND B) = 0.
P(A) * P(B) = [latex]\frac{3}{12}[/latex] * [latex]\frac{1}{12}[/latex] = [latex]\frac{3}{144}[/latex] .
P(A AND B) [latex]\ne[/latex] P(A)P(B), so A and B are dependent.
Try It
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.
 Compute P(T).
 Compute P(TF).
 Are T and F independent?.
 Are F and S mutually exclusive?
 Are F and S independent?
Show Answer
 P(T) = [latex]\frac{{1}}{{4}}[/latex]
 P(TF) = [latex]\frac{{1}}{{2}}[/latex]
 No
 No
 Yes
References
Lopez, Shane, Preety Sidhu. “U.S. Teachers Love Their Lives, but Struggle in the Workplace.” Gallup Wellbeing, 2013. http://www.gallup.com/poll/161516/teacherslovelivesstruggleworkplace.aspx (accessed May 2, 2013).
Data from Gallup. Available online at www.gallup.com/ (accessed May 2, 2013).
Concept Review
Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent.
In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other.
Formula Review
If A and B are independent, P(A AND B) = P(A)P(B), P(AB) = P(A) and P(BA) = P(B).
If A and B are mutually exclusive, P(A OR B) = P(A) + P(B) and P(A AND B) = 0.