Example

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 $1$ (ace), $2, 3, 4, 5, 6, 7, 8, 9, 10$, $J$ (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.

Example

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 $1$ (ace), $2, 3, 4, 5, 6, 7, 8, 9, 10$, $J$ (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 $QS$, $1D$, $1C$, $QD$.
2. Suppose you pick four cards and put each card back before you pick the next card. Your cards are $KH$, $7D$, $6D$, $KH$.

Which of $1$ or $2$ did you sample with replacement and which did you sample without replacement?

Example

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 of $A$, so $B = A'$. Also, $P(A) + P(B)$ = $P(A) + P(A') = 1$.
• The probabilities for $A$ and for $B$ are $P(A) = \displaystyle\frac{{3}}{{4}}$ and $P(B) = \displaystyle\frac{{1}}{{4}}$.
• Let $C$ = the event of getting all heads. $C$ = {$HH$}. Since $B = {TT}$, $P(B \text{ AND } C) = 0$. $B$ 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. $D = {TT}$. $P(D) = \displaystyle\frac{{1}}{{4}}$
• Let $E$ = event of getting a head on the first flip. (This implies you can get either a head or tail on the second flip.) $E = {HT,HH}$. $P(E) = \displaystyle\frac{{2}}{{4}} = \displaystyle\frac{{1}}{{2}}$
• 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.$F = {HT, TH, TT}$. $P(F) = \displaystyle\frac{{3}}{{4}}$

Example

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?

Example

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, $P(A) = \displaystyle\frac{{3}}{{6}}$ and $P(B) = \displaystyle\frac{{3}}{{6}}= \displaystyle\frac{{1}}{{2}}$.
• Let event $C$ = odd faces larger than two. Then $C$ = {$3, 5$}. Let event $D$ = all even faces smaller than five. Then $D$ = {$2, 4$}. $P(C \text{ AND } D) = 0$ 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. $E$ = {$1, 2, 3, 4$}.

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

• 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, $P(C|A) = \displaystyle\frac{{2}}{{3}}$.

Example

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 $P(G \text{ AND } H) = 0.3$. Are $G$ and $H$ independent?

If $G$ and $H$ are independent, then you must show ONE of the following:

• $P(G|H) = P(G)$
• $P(H|G) = P(H)$
• $P(G \text{ AND } H) = P(G)P(H)$

Note:

The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information.

1. Show that $P(G|H) = P(G)$.
2. Show $P(G \text{ AND } H) = P(G)P(H)$.

Example

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 $P(C \text{ AND } D) = 0.225$.

Justify your answers to the following questions numerically.

1. Are $C$ and $D$ independent?
2. Are $C$ and $D$ mutually exclusive?
3. What is $P(D|C)$?

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.

• $P(R) = \displaystyle\frac{{3}}{{8}}$.$P(B) = \displaystyle\frac{{5}}{{8}}$. $P(R \text{ AND } B) = 0$. (You cannot draw one card that is both red and blue.)
• $P(E) = \displaystyle\frac{{3}}{{8}}$. (There are three even-numbered cards, $R2, B2, \text{ and } B4$.)
• $P(E|B) = \displaystyle\frac{{2}}{{5}}$. (There are five blue cards: $B1, B2, B3, B4, \text{ and } B5$. Out of the blue cards, there are two even cards; $B2 \text{ and } B4$.)
• $P(B|E) = \displaystyle\frac{{2}}{{3}}$. (There are three even-numbered cards: $R2, B2, \text{ and } B4$. Out of the even-numbered cards, two are blue; $B2 \text{ and }B4$.)
• The events $R$ and $B$ are mutually exclusive because $P(R \text{ AND } B) = 0$.
• Let $G$ = card with a number greater than $3$. $G$ = {$B4, B5$}. $P(G) = \displaystyle\frac{{2}}{{8}}$. 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.

Example

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:
• $P(F) = 0.60$; $P(L) = 0.50$
• $P(F \text{ AND } L) = 0.45$
• $P(L|F) = 0.75$

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 $P(F|L)$ yet, so you cannot use the second condition.

Example

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 $P(A \text{ AND } B)$? If $P(A \text{ AND } B) = 0$, then $A$ and $B$ are mutually exclusive.)
8. Are $A$ and $B$ independent? (Hint: Is $P(A \text{ AND } B) = P(A)P(B)$? If $P(A \text{ AND } B) = P(A)P(B)$, then $A$ and $B$ are independent. If not, then they are dependent).