Independence and Mutual Exclusivity

Independence and Mutual Exclusivity

We briefly covered independent and mutually exclusive events but we will elaborate on them further in this section.

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 conditions below. 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.


This video provides a brief lesson on finding the probability of independent events.

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EXAMPLE 1

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.

  1. 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.

  2. 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.

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PRACTICE 1

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. Three cards are picked at random.

  1. 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? If so, which is it?

  2. 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? If so, which is it?

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PRACTICE 2

You have a fair, well-shuffled deck of 52 cards. It consist 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. S = spades, H = hearts, D = diamonds, C = clubs.

  1. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are QS, 1D, 1C, QD. Did you sample with or without replacement?

  2. Suppose you pick four cards and put each card back before you pick the next card. Your cards are KH, 7D, 6D, KH. Did you sample with or without replacement?

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PRACTICE 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.

  1. QS, 1D, 1C, QD

  2. KH, 7D, 6D, KH

  3. QS, 7D, 6D, KS

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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}.


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.

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PRACTICE 4

Draw two cards form a standard 52-card deck with replacement. Find the probability of getting at least one black card. (Hint: The sample space of drawing two cards with replacement from a standard 52-card deck with respect to color is {BB, BR, RB, RR}.)

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PRACTICE 5

Flip two fair coins. Find the probabilities of the events.

  1. Let F = the event of getting at most one tail (zero or one tail).

  2. Let G = the event of getting two faces that are the same.

  3. Let H = the event of getting a head on the first flip followed by a head or tail on the second flip.

  4. Are F and G mutually exclusive?

  5. Let J = the event of getting all tails. Are J and H mutually exclusive?

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This video provides two more examples of finding the probability of events that are mutually exclusive.


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PRACTICE 6

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). Find the probability of the following events:

  1. Let F = the event of getting the white ball twice.

  2. Let G = the event of getting two balls of different colors.

  3. Let H = the event of getting white on the first pick.

  4. Are F and G mutually exclusive?

  5. Are G and H mutually exclusive?

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PRACTICE 7

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}. 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) = 3/6

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(CAND 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}.

    1. Are C and E mutually exclusive events? (Answer yes or no.) Why or why not?

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PRACTICE 8

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 number 1,2,3 and 4.


R = a red marble

G = a green marble

O = an odd-numbered marble

  1. Determine the sample space

  2. What is P(G and O)?

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PRACTICE 9

Let event C = taking an English class. Let event D = taking a speech class.

Suppose P(C) = 0.75, P(D) = 0.3, P(C|D) = 0.75 and P(C AND D) = 0.225.

Justify your answers to the following questions numerically.

  1. Are C and D independent?

  2. Are C and D mutually exclusive?

  3. What is P(D|C)?

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PRACTICE 10

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.

  1. Find P(B|D).

  2. Find P(D|B).

  3. Are B and D independent?

  4. Are B and D mutually exclusive?

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EXAMPLE

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 = R1, R2, R3, B1, B2, B3, B4, B5. S has eight outcomes.

  • P(R) = 3/8. P(R AND B) = 0. (You cannot draw one card that is both red and blue.)

  • P(E) =3/8 (There are three even-numbered cards, R2, B2, and B4.)

  • P(E|B) = 2/5. (There are five blue cards: B1, B2, B3, B4, and B5. Out of the blue cards, there are two even cards; B2 andB4.)

  • P(B|E) = 2/3. (There are three even-numbered cards: R2, B2, and B4. Out of the even-numbered cards, to are blue; B2 andB4.)

  • The events R and B are mutually exclusive because P(R AND B) = 0.

  • Let G = card with a number greater than 3. G = {B4, B5}. P(G) = 2/8

  • P(G) = P(G|H), which means that G and H are independent.

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EXAMPLE

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?

P(B|A) = 0.67

P(B) = 0.25

So P(B) does not equal P(B|A) 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, becauseP(B AND A) = 0.20, not 0.

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PRACTICE 11

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) = 0.56

  1. What's the probability of P(I or F)?

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PRACTICE 12

  1. Toss one fair coin (the coin has two sides, H and T). The outcomes are ________. Count the outcomes. There are ____ outcomes.

  2. Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5 or 6 dots on a side). The outcomes are ________________. Count the outcomes. There are ___ outcomes.

  3. Multiply the two numbers of outcomes. The answer is _______.

  4. If you flip one fair coin and follow it with the toss of one fair, six-sided 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.)

  5. Event A = heads (H) on the coin followed by an even number (2, 4, 6) on the die.
    A = {_________________}. Find P(A).

  6. Event B = heads on the coin followed by a three on the die. B = {________}. Find P(B).

  7. 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.)

  8. 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).

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PRACTICE 13

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.

  1. Compute P(T).

  2. Compute P(T|F).

  3. Are T and F independent?.

  4. Are F and S mutually exclusive?

  5. Are F and S independent?

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References:

  1. https://courses.lumenlearning.com/introstats1/chapter/independent-and-mutually-exclusive-events/

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  • Ex: Probability of Events that are Mutually Exclusive Events . Authored by: Mathispower4u. Located at: https://youtu.be/zxhDOvS2c3k. License: All Rights Reserved. License Terms: Standard YouTube License

  • Probability of Independent Events. Authored by: Mathispower4u. Located at: https://youtu.be/QsnfXUqsFPU. License: All Rights Reserved. License Terms: Standard YouTube LIcense

Lopez, Shane, Preety Sidhu. “U.S. Teachers Love Their Lives, but Struggle in the Workplace.” Gallup Wellbeing, 2013. http://www.gallup.com/poll/161516/teachers-love-lives-struggle-workplace.aspx (accessed May 2, 2013).

Data from Gallup. Available online at www.gallup.com/ (accessed May 2, 2013).



Answers

Practice 1.

  1. Yes you can decide (because the Q of spades was picked twice). This is with replacement.

  2. No, you can't decide (nothing was chosen twice, so there is no clear evidence of replacement or not).

Practice 2.

  1. Without replacement.

  2. With replacement

Practice 3.

Without replacement:

  1. Possible

  2. Impossible

  3. Possible

With replacement:

  1. Possible

  2. Possible

  3. Possible

Practice 4.

Event A = getting at least one black card = {BB, BR, RB]

P(A) = 3/4 = 0.75

Practice 5.


  1. Zero (0) or one (1) tails occur when the outcomes HH, TH, HT show up. P(F)=3/4

  2. Two faces are the same if HH or TT show up. P(G) = 2/4

  3. 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) =2/4

  4. F and G share HH so P(F AND G) is not equal to zero (0). F and G are not mutually exclusive.

  5. 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.

PRACTICE 6.

  1. P(F) = 1/4

  2. P(G) = 1/2

  3. P(H) = 1/2

  4. Yes

  5. No

Practice 7.

  1. No. C = {3, 5} and E = {1, 2, 3, 4}. P(C AND E) = 1/6 To be mutually exclusive, P(C AND E) must be zero.

Practice 8.

  1. Space space = {R1,R2,R3,R4,R5,R6,G1,G2,G3,G4}

  2. Event of (G and O) = {G1, G3}. P(G and O) = 2/10 = 0.20

Practice 9

  1. Yes, because P(C|D) = P(C).

  2. No, because P(C AND D) is not equal to zero.

  3. P(DmidC) = P(C and D)/P(C)=0.225/0.75=0.3

Practice 10.

  1. P(B|D) = 0.6667

  2. P(D|B) = 0.5

  3. No

  4. No

Practice 11.

  1. Conceptually, Mark only has two options. He can take the interstate or fifth street; he can't take both. There are no other ways to get to work. Therefore, the probability of him taking either the interstate of fifth street is 1 (he has to take one of these routes to get to work).

Mathematically, P(I and F) = 0 and P(I or F) = P(I) + P(F) - P(I and F) = 0.44 + 0.56 + 0 = 1.

Practice 12.

  1. H and T; 2

  2. 1, 2, 3, 4, 5, 6; 6

  3. 2(6) = 12

  4. T1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6

  5. A = {H2, H4, H6}; P(A) = 3/12

  6. B = {H3}; P(B) = 1/12

  7. Yes, because P(A AND B) = 0

  8. P(A AND B) = 0. P(A)P(B) = (3/12)(1/12). P(A AND B) does not equal P(A)P(B), so A and B are dependent.

Practice 13.

  1. P(T) = 1/4

  2. P(T|F) = 1/2

  3. No

  4. No

  5. Yes