Independent and mutually exclusive do **not** mean the same thing.

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### 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 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, 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.

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 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.

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.

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.

*Q*of spades,

*K*of hearts and

*Q*of spades. Can you decide if the sampling was with or without 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?

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.

*QS*, 1

*D*, 1

*C*,

*QD*.Suppose you pick four cards and put each card back before you pick the next card. Your cards are

*KH*, 7

*D*, 6

*D*,

*KH*.

Which of a. or b. did you sample with replacement and which did you sample 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, *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.

*QS*, 1

*D*, 1

*C*,

*QD*

*KH*, 7

*D*, 6

*D*,

*KH*

*QS*, 7

*D*, 6

*D*,

*KS*

### 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*) =

*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.

*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′*. Also,

*P*(

*A*) +

*P*(

*B*) =

*P*(

*A*) +

*P*(

*A′*) = 1.The probabilities for

*A*and for

*B*are

*P*(

*A*) = and

*P*(

*B*) = .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*) = 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*) = 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*) =

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

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?

Look at the sample space in (Figure).

Zero (0) or one (1) tails occur when the outcomes*HH*,

*TH*,

*HT*show up.

*P*(

*F*) = Two faces are the same if

*HH*or

*TT*show up.

*P*(

*G*) = 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*) =

*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.

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

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

*H*= the event of getting white on the first pick.Are

*F*and

*G*mutually exclusive?Are

*G*and

*H*mutually exclusive?

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}.

*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*) = and

*P*(

*B*) = .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}.

No. *C* = {3, 5} and *E* = {1, 2, 3, 4}. *P*(*C* AND *E*) =

*P*(

*C*AND

*E*) must be zero.

Find

*P*(

*C*|

*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*|

*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*) = .

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:

*P*(

*A*|

*B*) =

*P*(

*A*)

*P*(

*B*|

*A*) =

*P*(

*B*)

*P*(

*A*AND

*B*) =

*P*(

*A*)

*P*(

*B*)

*P*(*A*|*B*) =

The events are independent because *P*(*A*|*B*) = *P*(*A*).

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?

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*)

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