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4.3: Independent and Mutually Exclusive Events

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    Independent and mutually exclusive do not mean the same thing.

    Independent Events

    Two events are independent if the following are true:

    • \(P(A \mid B)=P(A)\)
    • \(P(B \mid A)=P(B)\)
    • \(P(A \cap B)=P(A) P(B)\)

    Two events A and B are independent events if the knowledge that one occurred does not affect the chance the other occurs. For example, if we measure the blood pressure of two unrelated participants in a community health fair, these are independent events. The outcome of the first participant's screening does not change the probability for the outcome of the second participant's screening. 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. For instance, if a researcher is randomly selecting electronic health records to audit but allows the same record to be potentially selected again for a follow-up review, then events are considered to be independent. This means 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. This is the standard practice in clinical trials; once a participant is assigned to an exercise intervention group, they are no longer available to be picked for the control group. 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.
    Exercise \(\PageIndex{1}\)

    You have a pool of 52 participants in a community nutrition study. It consists of four different regional health centers (North, South, East, and West). There are 13 participants from each center, assigned ID numbers 1 through 13.

    a. Sampling with replacement: Suppose you randomly select three participant IDs with replacement to conduct a survey satisfaction audit. The first ID you pick out of the 52 is Participant #12 from the North Center. You record the data, put this ID back into the digital pool, and pick a second ID from the 52-person list. It is Participant #5 from the West Center. You put this ID back and pick a third ID. This time, the ID is Participant #12 from the North Center again. Your picks are {North #12, West #5, North #12}. You have picked the same participant twice. You pick each ID from the original 52-person pool.

    b. Sampling without replacement: Suppose you pick three participants to receive a free wearable fitness tracker without replacement. The first participant you pick out of the 52 is Participant #13 from the East Center. You assign them the tracker and pick the second participant from the 51 remaining in the pool. It is Participant #3 from the South Center. You assign them the tracker and pick the third participant from the remaining 50 people. The third participant is Participant #11 from the West Center. Your picks are {East #13, South #3, West #11}. Because you have picked the participants without replacement, you cannot pick the same person twice.

    Exercise \(\PageIndex{2}\)

    You have a pool of 52 participants in a community nutrition study. It consists of four regional health centers: North (N), South (S), East (E), and West (W). There are 13 participants from each center, assigned ID numbers 1 through 13.

    a. Suppose you pick four participant IDs, but do not put any IDs back into the pool. Your selected IDs are {12-W, 1-E, 1-N, 12-E}.

    b. Suppose you pick four participant IDs and put each ID back into the pool before you pick the next one. Your selected IDs are {13-S, 7-E, 6-E, 13-S}.

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

    Answer

    a. Without replacement; b. With replacement

    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 \(P(A \cap B)=0\).

    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 \(C=\{7,9\}\). \(A \cap B=\{4,5\} . P(A \cap B)=\dfrac{2}{10}\) 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 \(P(A \cap C)=0\). 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.

    Exercise \(\PageIndex{3}\)

    Flip two fair coins. (This is an experiment.)

    The sample space is \(\{H H, H T, T H, T T\}\) where \(T=\) tails and \(H=\) heads. The outcomes are \(H H, H T, T H\), and \(T T\). The outcomes HT and TH are different. The HT means that the first coin showed heads and the second coin showed tails. The \(T H\) 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 \(\{H H, H T, T H\}\). The outcome \(H H\) shows zero tails. \(H T\) and \(T H\) each show one tail.
    • Let \(B=\) the event of getting all tails. \(B\) can be written as \(\{T T\}\). \(B\) is the complement of \(A\), so \(B=A^{\prime}\). Also, \(P(A)+P(B)=P(A)+P(A)=1\).
    • The probabilities for \(A\) and for \(B\) are \(P(A)=\dfrac{3}{4}\) and \(P(B)=\dfrac{1}{4}\).
    • Let \(C=\) the event of getting all heads. \(C=\{H H\}\). Since \(B=\{T T\}, P(B \cap C)=0\). \(B\) 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. \(D=\{T T\} . P(D)=\dfrac{1}{4}\)
    • 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.) \(E=\{H T, H H\} . P(E)=\dfrac{2}{4}\)
    • 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. \(F=\{H T, T H, T T\} . P(F)=\dfrac{3}{4}\)

    Exercise \(\PageIndex{4}\)

    Perform two different balance assessments on a patient: the Single Leg Stance and the Tandem Walk. Each test is recorded as a Pass (P) or a Fail (F). Assume the probability of passing each is equal (0.5). The sample space is {PP, PF, FP, FF}.

    Let F = the event of failing at most one test (zero or one fail).

    Let G = the event of getting the same result on both tests.

    Let H = the event of passing the first test followed by a pass or fail on the second test.

    1. Are F and G mutually exclusive?

    2. Let J = the event of failing both tests. Are J and H mutually exclusive?

    Answer

    Look at the sample space: {PP, PF, FP, FF}.

    a. Zero (0) or one (1) fails occur when the outcomes {PP, FP, PF} show up. P(F)=43​

    b. Two results are the same if {PP} or {FF} show up. P(G)=42​

    c. A pass on the first test followed by a pass or fail on the second occurs when {PP} or {PF} show up. P(H)=42​

    d. F and G share the outcome {PP}, so P(F∩G) is not equal to zero. F and G are not mutually exclusive.

    e. Failing both tests occurs when {FF} shows up. H's outcomes are {PP} and {PF}. J and H have nothing in common, so P(J∩H)=0. J and H are mutually exclusive.

    Exercise \(\PageIndex{5}\)

    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 \(A, A^{\prime}\). The complement of \(A, A^{\prime}\), is \(B\) because \(A\) and \(B\) together make up the sample space. \(P(A)+P(B)=P(A)+P(A)=1\). Also, \(P(A)=\dfrac{3}{6}\) and \(P(B)=\dfrac{3}{6}\).
    • Let event \(C=\) odd faces larger than two. Then \(C=\{3,5\}\). Let event \(D=\) all even faces smaller than five. Then \(D=\{2,4\} . P(C \cap D)=0\) 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?

    Answer
    • Find \(P(C \mid A)\). This is a conditional probability. Recall that the event \(C\) is \(\{3,5\}\) and event \(A\) is \(\{1,3\), 5\}. To find \(P(C \mid A)\), 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, \(P(C \mid A)=\dfrac{2}{3}\).
    Exercise \(\PageIndex{7}\)

    Let event C = taking a nutrition course. Let event D = participating in a physical activity program. 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)?

    Answer

    a. Yes, because P(C∣D)=P(C). (The fact that a student participates in a physical activity program does not change the probability that they are also taking a nutrition course.)

    b. No, because P(C∩D) is not equal to zero. (There is a 22.5% overlap where students are doing both.)

    c. P(D∣C)=P(C)P(C∩D)​=0.750.225​=0.3

    Exercise \(\PageIndex{9}\)

    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 \(F\) 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:

    • \(P(F)=0.60 ; P(L)=0.50\)
    • \(P(F \cap L)=0.45\)
    • \(P(L \mid F)=0.75\)
    Note

    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 \(P(F \mid L)\) yet, so you cannot use the second condition.

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

    Check whether \(P(L \mid F)\) equals \(P(L)\). We are given that \(P(L \mid F)=0.75\), but \(P(L)=0.50\); 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.

    This page titled 4.3: Independent and Mutually Exclusive Events is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.