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Independent and mutually exclusive do not mean the same thing.

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Independent EventsMutually Exclusive EventsFormula ReviewBringing It TogetherWhen two events are mutually exclusive the probability of either one of those occurring is the sum of individual probabilities?What is the joint probability of two mutually exclusive events give one example?When two events are mutually exclusive events?

Independent Events

Two events are independent if the following are true:

(P(textA) = P(textA))(P(textB) = P(textB))(P(textA AND B) = P(textA)P(textB))

Two events (textA) and (textB) 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 a population

Sampling may be done with replacement or without replacement (Figure (PageIndex1)):

With replacement:

Without replacement:

Figure (PageIndex1): A visual representation of the sampling process. If the sample items are replaced after each sampling sự kiện, then this is "sampling with replacement" if not, then it is "sampling without replacement". Image used with permission (CC BY-SA 4.0; Dan Kernler).

If it is not known whether (textA) and (textB) are independent or dependent, assume they are dependent until you can show otherwise.

Example (PageIndex1): Sampling with and without replacement

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, (textJ) (jack), (textQ.) (queen), (textK) (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 (textQ.) 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 (textQ.) of spades again. Your picks are (textQ.) of spades, ten of clubs, (textQ.) of spades. You have picked the (textQ.) of spades twice. You pick each card from the 52-card deck.

b. Sampling without replacement:

Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is the (textK) 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 (textJ) of spades. Your picks are (textK) of hearts, three of diamonds, (textJ) of spades. Because you have picked the cards without replacement, you cannot pick the same card twice.

Exercise (PageIndex1)

Suppose you know that the picked cards are (textQ.) of spades, (textK) of hearts and (textQ.)of spades. Can you decide if the sampling was with or without replacement?Suppose you know that the picked cards are (textQ.) of spades, (textK) of hearts, and (textJ)of spades. Can you decide if the sampling was with or without replacement?Answer a

With replacement

Answer b

Example (PageIndex2)

Suppose you pick four cards, but do not put any cards back into the deck. Your cards are (textQS, 1textD, 1textC, textQD).Suppose you pick four cards and put each card back before you pick the next card. Your cards are (textKH, 7textD, 6textD, textKH).

Which of a. or b. did you sample with replacement and which did you sample without replacement?

Answer a

Without replacement

Answer b

With replacement

Exercise (PageIndex2)

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, (textJ) (jack), (textQ.) (queen), and (textK) (king) of that suit. (textS =) spades, (textH =) Hearts, (textD =) Diamonds, (textC =) Clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement.

(textQS, 1textD, 1textC, textQD)(textKH, 7textD, 6textD, textKH)(textQS, 7textD, 6textD, textKS)Answer - without replacement

a. Possible; b. Impossible, c. Possible

Answer - with replacement

a. Possible; c. Possible, c. Possible

Mutually Exclusive Events

(textA) and (textB) are mutually exclusive events if they cannot occur the same time. This means that (textA) and (textB) do not share any outcomes and (P(textA AND B) = 0).

For example, suppose the sample space

[S = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. nonumber]

Let (textA = 1, 2, 3, 4, 5, textB = 4, 5, 6, 7, 8\), and (textC = 7, 9\). (textA AND B = 4, 5\).

[P(textA AND B) = dfrac210 nonumber]

and is not equal to zero. Therefore, (textA) and (textB) are not mutually exclusive. (textA) and (textC) do not have any numbers in common so (P(textA AND C) = 0). Therefore, (textA) and (textC) are mutually exclusive.

If it is not known whether (textA) and (textB) are mutually exclusive, assume they are not until you can show otherwise. The following examples illustrate these definitions and terms.

Example (PageIndex3)

Flip two fair coins.

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.

most one tail

complement

more than one

least one

Exercise (PageIndex3)

Draw two cards from a standard 52-card deck with replacement. Find the probability of getting least one black card.

Answer

The sample space of drawing two cards with replacement from a standard 52-card deck with respect to color is (BB, BR, RB, RR\).

Event (A =) Getting least one black card (= BB, BR, RB\)

(P(textA) = dfrac34 = 0.75)

Example (PageIndex4)

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

Let (textF =) the sự kiện of getting most one tail (zero or one tail).Let (textG =) the sự kiện of getting two faces that are the same.Let (textH =) the sự kiện of getting a head on the first flip followed by a head or tail on the second flip.Are (textF) and (textG) mutually exclusive?Let (textJ =) the sự kiện of getting all tails. Are (textJ) and (textH) mutually exclusive?

Solution

Look the sample space in Example (PageIndex3).

Zero (0) or one (1) tails occur when the outcomes (HH, TH, HT) show up. (P(textF) = dfrac34)Two faces are the same if (HH) or (TT) show up. (P(textG) = dfrac24)A head on the first flip followed by a head or tail on the second flip occurs when (HH) or (HT) show up. (P(textH) = dfrac24)(textF) and (textG) share (HH) so (P(textF AND G)) is not equal to zero (0). (textF) and (textG) are not mutually exclusive.Getting all tails occurs when tails shows up on both coins ((TT)). (textH)’s outcomes are (HH) and (HT).

(textJ) and (textH) have nothing in common so (P(textJ AND H) = 0). (textJ) and (textH) are mutually exclusive.

Exercise (PageIndex4)

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:

Let (textF =) the sự kiện of getting the white ball twice.Let (textG =) the sự kiện of getting two balls of different colors.Let (textH =) the sự kiện of getting white on the first pick.Are (textF) and (textG) mutually exclusive?Are (textG) and (textH) mutually exclusive?

Answer

(P(textF) = dfrac14)(P(textG) = dfrac12)(P(textH) = dfrac12)YesNo

Example (PageIndex5)

Roll one fair, six-sided die. The sample space is 1, 2, 3, 4, 5, 6. Let sự kiện (textA =) a face is odd. Then (textA = 1, 3, 5\). Let sự kiện (textB =) a face is even. Then (textB = 2, 4, 6\).

Find the complement of (textA), (textA′). The complement of (textA), (textA′), is (textB) because (textA) and (textB) together make up the sample space. (P(textA) + P(textB) = P(textA) + P(textA′) = 1). Also, (P(textA) = dfrac36) and (P(textB) = dfrac36).Let sự kiện (textC =) odd faces larger than two. Then (textC = 3, 5\). Let sự kiện (textD =) all even faces smaller than five. Then (textD = 2, 4\). (P(textC AND D) = 0) because you cannot have an odd and even face the same time. Therefore, (textC) and (textD) are mutually exclusive events.Let sự kiện (textE =) all faces less than five. (textE = 1, 2, 3, 4\).

Are (textC) and (textE) mutually exclusive events? (Answer yes or no.) Why or why not?

Answer

No. (textC = 3, 5\) and (textE = 1, 2, 3, 4\). (P(textC AND E) = dfrac16). To be mutually exclusive, (P(textC AND E)) must be zero.

Find (P(textA)). This is a conditional probability. Recall that the sự kiện (textC) is 3, 5 and sự kiện (textA) is 1, 3, 5. To find (P(textA)), find the probability of (textC) using the sample space (textA). You have reduced the sample space from the original sample space 1, 2, 3, 4, 5, 6 to 1, 3, 5. So, (P(textA) = dfrac23).

Exercise (PageIndex5)

Let sự kiện (textA =) learning Spanish. Let sự kiện (textB) = learning German. Then (textA AND B) = learning Spanish and German. Suppose (P(textA) = 0.4) and (P(textB) = 0.2). (P(textA AND B) = 0.08). Are events (textA) and (textB) independent? Hint: You must show ONE of the following:

(P(textA) = P(textA))(P(textA))(P(textA AND B) = P(textA)P(textB))

Answer

[P(textB) = dfractextP(A AND B)P(textB) = dfrac0.080.2 = 0.4 = P(textA)]

The events are independent because (P(textB) = P(textA)).

Example (PageIndex6)

Let sự kiện (textG =) taking a math class. Let sự kiện (textH =) taking a science class. Then, (textG AND H =) taking a math class and a science class. Suppose (P(textG) = 0.6), (P(textH) = 0.5), and (P(textG AND H) = 0.3). Are (textG) and (textH) independent?

If (textG) and (textH) are independent, then you must show ONE of the following:

(P(textH) = P(textG))(P(textG) = P(textH))(P(textG AND H) = P(textG)P(textH))

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(textG) = P(textG)).b. Show (P(textG AND H) = P(textG)P(textH)).

Solution

(P(textH) = dfracP(textG AND H)P(textH) = dfrac0.30.5 = 0.6 = P(textG))(P(textG)P(textH) = (0.6)(0.5) = 0.3 = P(textG AND H))

Since (textG) and (textH) 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. For practice, show that (P(textH) = P(textH)) to show that (textG) and (textH) are independent events.

Exercise (PageIndex6)

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.

(textR =) a red marble(textG =) a green marble(textO =) an odd-numbered marbleThe sample space is (textS = R1, R2, R3, R4, R5, R6, G1, G2, G3, G4\).

(textS) has ten outcomes. What is (P(textG AND O))?

Answer

Event (textG) and (textO = G1, G3\)

(P(textG and O) = dfrac210 = 0.2)

Example (PageIndex7)

Let sự kiện (textC =) taking an English class. Let sự kiện (textD =) taking a speech class.

Suppose (P(textC) = 0.75), (P(textD) = 0.3), (P(textD) = 0.75) and (P(textC AND D) = 0.225).

Justify your answers to the following questions numerically.

Are (textC) and (textD) independent?Are (textC) and (textD) mutually exclusive?What is (P(textD))?

Solution

Yes, because (P(textC) = P(textC)).No, because (P(textC AND D)) is not equal to zero.(P(textD) = dfracP(textC AND D)P(textC) = dfrac0.2250.75 = 0.3)

Exercise (PageIndex7)

A student goes to the library. Let events (textB =) the student checks out a book and (textD =) the student checks out a DVD. Suppose that (P(textB) = 0.40), (P(textD) = 0.30) and (P(textB AND D) = 0.20).

Find (P(textD)).Find (P(textD)).Are (textB) and (textD) independent?Are (textB) and (textD) mutually exclusive?

Answer

(P(textB) = 0.6667)(P(textD) = 0.5)NoNo

Example (PageIndex8)

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

(textR =) red card is drawn,(textB =) blue card is drawn,(textE =) even-numbered card is drawn.

The sample space (S = R1, R2, R3, B1, B2, B3, B4, B5).

(S) has eight outcomes.

(P(textR) = dfrac38). (P(textB) = dfrac58). (P(textR AND B) = 0). (You cannot draw one card that is both red and blue.)(P(textE) = dfrac38). (There are three even-numbered cards, (R2, B2), and (B4).)(P(textB) = dfrac25). (There are five blue cards: (B1, B2, B3, B4), and (B5). Out of the blue cards, there are two even cards; (B2) and (B4).)(P(textB) = dfrac23). (There are three even-numbered cards: (R2, B2), and (B4). Out of the even-numbered cards, to are blue; (B2) and (B4).)The events (textR) and (textB) are mutually exclusive because (P(textR AND B) = 0).Let (textG =) card with a number greater than 3. (textG = B4, B5\). (P(textG) = dfrac28). Let (textH =) blue card numbered between one and four, inclusive. (textH = B1, B2, B3, B4\). (P(textG) = frac14). (The only card in (textH) that has a number greater than three is B4.) Since (dfrac28 = dfrac14), (P(textG) = P(textH)), which means that (textG) and (textH) are independent.

Exercise (PageIndex8)

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 (textA) be the sự kiện that a fan is rooting for the away team.

Let (textB) be the sự kiện that a fan is wearing blue.

Are the events of rooting for the away team and wearing blue independent? Are they mutually exclusive?

Answer

(P(textA) = 0.67)(P(textB) = 0.25)

So (P(textB)) does not equal (P(textA)) which means that (textB and textA) are not independent (wearing blue and rooting for the away team are not independent). They are also not mutually exclusive, because (P(textB AND A) = 0.20), not (0).

Example (PageIndex9)

In a particular college class, 60% of the students are female. Fifty percent of all students in the class have long hair. Forty-five percent of the students are female and have long hair. Of the female students, 75% have long hair. Let (textF) be the sự kiện that a student is female. Let (textL) be the sự kiện 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(textF) = 0.60); (P(textL) = 0.50)(P(textF AND L) = 0.45)(P(textF) = 0.75)

The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know (P(textF)) yet, so you cannot use the second condition.

Solution 1

Check whether (P(textF AND L) = P(textF)P(textL)). We are given that (P(textF AND L) = 0.45), but (P(textF)P(textL) = (0.60)(0.50) = 0.30). The events of being female and having long hair are not independent because (P(textF AND L)) does not equal (P(textF)P(textL)).

Solution 2

Check whether (P(textL)) equals (P(textL)). We are given that (P(textL) = 0.75), but (P(textL) = 0.50); they are not equal. The events of being female and having long hair are not independent.

Interpretation of Results

The 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.

Exercise (PageIndex9)

Mark is deciding which route to take to work. His choices are (textI = textthe Interstate) and (textF = textFifth Street)

(P(textI) = 0.44) and (P(textF) = 0.55)(P(textI AND F) = 0) because Mark will take only one route to work.

What is the probability of (P(textI OR F))?

Answer

Because (P(textI AND F) = 0),

(P(textI OR F) = P(textI) + P(textF) - P(textI AND F) = 0.44 + 0.56 - 0 = 1)

Example (PageIndex10)

Toss one fair coin (the coin has two sides, (textH) and (textT)). The outcomes are ________. Count the outcomes. There are ____ outcomes.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.Multiply the two numbers of outcomes. The answer is _______.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).)Event (textA =) heads ((textH)) on the coin followed by an even number (2, 4, 6) on the die.
(textA) = _________________. Find (P(textA)).Event (textB =) heads on the coin followed by a three on the die. (textB =) ________. Find (P(textB)).Are (textA) and (textB) mutually exclusive? (Hint: What is (P(textA AND B))? If (P(textA AND B) = 0), then (textA) and (textB) are mutually exclusive.)Are (textA) and (textB) independent? (Hint: Is (P(textA AND B) = P(textA)P(textB))? If (P(textA AND B) = P(textA)P(textB)), then (textA) and (textB) are independent. If not, then they are dependent).

Solution

(textH) and (textT); 21, 2, 3, 4, 5, 6; 62(6) = 12(T1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6)(textA = H2, H4, H6\); (P(textA) = dfrac312)(textB = H3\); (P(textB) = dfrac112)Yes, because (P(textA AND B) = 0)(P(textA AND B) = 0). (P(textA)P(textB) = left(dfrac312right)left(dfrac112right)). (P(textA AND B)) does not equal (P(textA)P(textB)), so (textA) and (textB) are dependent.

Exercise (PageIndex10)

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 (textT) be the sự kiện of getting the white ball twice, (textF) the sự kiện of picking the white ball first, (textS) the sự kiện of picking the white ball in the second drawing.

Compute (P(textT)).Compute (P(textT)).Are (textT) and (textF) independent?.Are (textF) and (textS) mutually exclusive?Are (textF) and (textS) independent?

Answer

(P(textT) = dfrac14)(P(textT) = dfrac12)NoNoYes

References

Lopez, Shane, Preety Sidhu. “U.S. Teachers Love Their Lives, but Struggle in the Workplace.” Gallup Wellbeing, 2013. ://www.gallup.com/poll/161516/te...workplace.aspx (accessed May 2, 2013).Data from Gallup. Available online www.gallup.com/ (accessed May 2, 2013).

Review

Two events (textA) and (textB) 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 thành viên of a population is replaced after it is picked, so that thành viên has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each thành viên 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 (textA) and (textB) are independent, (P(textA AND B) = P(textA)P(textB), P(textA) = P(textA)) and (P(textA) = P(textB)).If (textA) and (textB) are mutually exclusive, (P(textA OR B) = P(textA) + P(textB) and P(textA AND B) = 0).

Exercise (PageIndex11)

(textE) and (textF) are mutually exclusive events. (P(textE) = 0.4); (P(textF) = 0.5). Find (P(textE∣F)).

Exercise (PageIndex12)

(textJ) and (textK) are independent events. (P(textJ) = 0.3). Find (P(textJ)).

Answer

(P(textJ) = 0.3)

Exercise (PageIndex13)

(textU) and (textV) are mutually exclusive events. (P(textU) = 0.26); (P(textV) = 0.37). Find:

(P(textU AND V) =)(P(textU) =)(P(textU OR V) =)

Exercise (PageIndex14)

(textQ.) and (textR) are independent events. (P(textQ.) = 0.4) and (P(textQ. AND R) = 0.1). Find (P(textR)).

Answer

(P(textQ. AND R) = P(textQ.)P(textR))

(0.1 = (0.4)P(textR))

(P(textR) = 0.25)

Bringing It Together

Exercise (PageIndex16)

A previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data are compiled into Table.

Shirt#≤ 210211–250251–290 290≤1–33 21 5 0 0 34–66 6 18 7 4 66–99 6 12 22 5

For the following, suppose that you randomly select one player from the 49ers or Cowboys.

If having a shirt number from one to 33 and weighing most 210 pounds were independent events, then what should be true about (P(textShirt #1–33|leq 210 text pounds))?

Exercise (PageIndex17)

The probability that a male develops some form of cancer in his lifetime is 0.4567. The probability that a male has least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Some of the following questions do not have enough information for you to answer them. Write “not enough information” for those answers. Let (textC =) a man develops cancer in his lifetime and (textP =) man has least one false positive.

(P(textC) =) ______(P(textC) =) ______(P(textC') =) ______If a test comes up positive, based upon numerical values, can you assume that man has cancer? Justify numerically and explain why or why not.

Answer

(P(textC) = 0.4567)not enough informationnot enough informationNo, because over half (0.51) of men have least one false positive text

Exercise (PageIndex18)

Given events (textG) and (textH: P(textG) = 0.43); (P(textH) = 0.26); (P(textH AND G) = 0.14)

Find (P(textH OR G)).Find the probability of the complement of sự kiện ((textH AND G)).Find the probability of the complement of sự kiện ((textH OR G)).

Exercise (PageIndex19)

Given events (textJ) and (textK: P(textJ) = 0.18); (P(textK) = 0.37); (P(textJ OR K) = 0.45)

Find (P(textJ AND K)).Find the probability of the complement of sự kiện ((textJ AND K)).Find the probability of the complement of sự kiện ((textJ AND K)).

Answer

(P(textJ OR K) = P(textJ) + P(textK) − P(textJ AND K); 0.45 = 0.18 + 0.37 - P(textJ AND K)); solve to find (P(textJ AND K) = 0.10)(P(textNOT (J AND K)) = 1 - P(textJ AND K) = 1 - 0.10 = 0.90)(P(textNOT (J OR K)) = 1 - P(textJ OR K) = 1 - 0.45 = 0.55)

Glossary

Dependent EventsIf two events are NOT independent, then we say that they are dependent.Sampling with ReplacementIf each thành viên of a population is replaced after it is picked, then that thành viên has the possibility of being chosen more than once.Sampling without ReplacementWhen sampling is done without replacement, each thành viên of a population may be chosen only once.The Conditional Probability of One Event Given Another EventP(A|B) is the probability that sự kiện A will occur given that the sự kiện B has already occurred.The OR of Two EventsAn outcome is in the sự kiện A OR B if the outcome is in A, is in B, or is in both A and B.

When two events are mutually exclusive the probability of either one of those occurring is the sum of individual probabilities?

The addition rule for probabilities describes two formulas, one for the probability for either of two mutually exclusive events happening and the other for the probability of two non-mutually exclusive events happening. The first formula is just the sum of the probabilities of the two events.

What is the joint probability of two mutually exclusive events give one example?

Probability Rules for Mutually Exclusive Events Since the events cannot occur simultaneously, their joint probability is always zero.

When two events are mutually exclusive events?

If two events have no elements in common (Their intersection is the empty set.), the events are called mutually exclusive. Thus, P(A∩B)=0 . This means that the probability of sự kiện A and sự kiện B happening is zero. They cannot both happen. Tải thêm tài liệu liên quan đến nội dung bài viết If two events are mutually exclusive, what is the probability that one or the other occurs