Independent and Mutually Exclusive Events

Learning objectives

Independent and mutually exclusive do not mean the same thing.

Independent Events

Two events are independent if the following are true:

• P(A|B) = P(A)
• P(B|A) = P(B)
• P(A AND 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, 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 events.

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.

A bag contains four blue and three white marbles. James draws one marble from the bag at random, records the color, and replaces the marble. The probability of drawing blue is $\frac{4}{7}$. When James draws a marble from the bag a second time, the probability of drawing blue is still $\frac{4}{7}$. James replaced the marble after the first draw, so there are still four blue and three white marbles.

• 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.

The bag still contains four blue and three white marbles. Maria draws one marble from the bag at random, records the color, and sets the marble aside. The probability of drawing blue on the first draw is $\frac{4}{7}$. Suppose Maria draws a blue marble and sets it aside. When she draws a marble from the bag a second time, there are now three blue and three white marbles. So, the probability of drawing blue is now $\frac{3}{6}=\frac{1}{2}$. Removing the first marble without replacing it influences the probabilities on the second draw.

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, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. Clubs and spades are black, while diamonds and hearts are red cards. There are 13 cards in each suit consisting of A (ace), 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 10 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, 10 of clubs, Q of spades}. You have picked the Q 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 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.

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}. P(A AND B) = $\frac{2}{10}$ 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 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.

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 event of A, so B = A′. Also, P(A) + P(B) = P(A) + P(A′) = 1.
• The probabilities for A and for B are P(A) = $\frac{3}{4}$ and P(B) = $\frac{1}{4}$.
• Let C = the event of getting all heads. C = {HH}. 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}. P(D) = $\frac{1}{4}\text{.}$
• 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) = $\frac{2}{4}\text{.}$
• 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) = $\frac{3}{4}\text{.}$

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) = $\frac{3}{6}$ and P(B) = $\frac{3}{6}$.
• 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 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. E = {1, 2, 3, 4}.

• Find P(C|A). This is a conditional probability. Recall that 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) = $\frac{2}{3}$.

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) = .6, P(H) = .5, and P(G AND H) = .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 AND H) = P(G)P(H)

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. For practice, show that P(H|G) = P(H) to show that G and H are independent events.

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) = $\frac{3}{8}$. P(B) = $\frac{5}{8}$. P(R AND B) = 0. You cannot draw one card that is both red and blue.
• P(E) = $\frac{3}{8}$. There are three even-numbered cards, R2, B2, and B4.
• P(E|B = $\frac{2}{5}$. 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(B|E) = $\frac{2}{3}$. There are three even-numbered cards: R2, B2, and B4. Out of the even-numbered cards, two are blue; B2 and B4.
• 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) = $\frac{2}{8}$. Let H = blue card numbered between one and four, inclusive. H = {B1, B2, B3, B4}. P(G|H) = $\frac{1}{4}$. The only card in H that has a number greater than three is B4. Since $\frac{2}{8}$ = $\frac{1}{4}$, P(G) = P(G|H), which means that G and H are independent.

In a particular class, 60 percent 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 percent 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(L|F) = 0.75

Solution 1

Solution 2

Interpretation of Results

References

Gallup. (n.d.). Retrieved from www.gallup.com/

Lopez, S., and Sidhu, P. (2013, March 28). U.S. teachers love their lives, but struggle in the workplace. Gallup Wellbeing. http://www.gallup.com/poll/161516/teachers-love-lives-struggle-workplace.aspx

Chapter 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(A|B) = P(A), and P(B|A) = P(B).

If A and B are mutually exclusive, P(A OR B) = P(A) + P(B) and P(A AND B) = 0.

Homework

Use the following information to answer the next 12 exercises. The graph shown is based on more than 170,000 interviews that took place from January through December 2012. The sample consists of employed Americans 18 years of age or older. The Health Index Scores are the sample space. We randomly sample one type of Health Index Score, the emotional well-being score.

Bringing It Together

Glossary

dependent events

if two events are NOT independent, then we say that they are dependent

sampling 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

sampling without replacement

when sampling is done without replacement, each member of a population may be chosen only once

the conditional probability of one event GIVEN another event

P(A|B) is the probability that event A will occur given that the event B has already occurred

the OR of two events

an outcome is in the event A OR B if the outcome is in A, is in B, or is in both A and B