Example

You have a fair, well-shuffled deck of [latex]52[/latex] cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are [latex]13[/latex] cards in each suit consisting of [latex]1, 2, 3, 4, 5, 6, 7, 8, 9, 10[/latex], [latex]J[/latex] (jack), [latex]Q[/latex] (queen), [latex]K[/latex] (king) of that suit.

1. Sampling with replacement:Suppose you pick three cards with replacement. The first card you pick out of the [latex]52[/latex] cards is the [latex]Q[/latex] of spades. You put this card back, reshuffle the cards and pick a second card from the [latex]52[/latex]-card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the [latex]52[/latex]-card deck. This time, the card is the [latex]Q[/latex] of spades again. Your picks are {[latex]Q[/latex] of spades, ten of clubs, [latex]Q[/latex] of spades}. You have picked the [latex]Q[/latex] of spades twice. You pick each card from the [latex]52[/latex]-card deck.
2. Sampling without replacement:Suppose you pick three cards without replacement. The first card you pick out of the [latex]52[/latex] cards is the [latex]K[/latex] of hearts. You put this card aside and pick the second card from the [latex]51[/latex] cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining [latex]50[/latex] cards in the deck. The third card is the [latex]J[/latex] of spades. Your picks are {[latex]K[/latex] of hearts, three of diamonds, [latex]J[/latex] 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 [latex]1, 2, 3, 4, 5, 6, 7, 8, 9, 10[/latex], [latex]J[/latex] (jack), [latex]Q[/latex] (queen), [latex]K[/latex] (king) of that suit. [latex]S[/latex] = spades, [latex]H[/latex] = Hearts, [latex]D[/latex] = Diamonds, [latex]C[/latex] = Clubs.

1. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are [latex]QS[/latex], [latex]1D[/latex], [latex]1C[/latex], [latex]QD[/latex].
2. Suppose you pick four cards and put each card back before you pick the next card. Your cards are [latex]KH[/latex], [latex]7D[/latex], [latex]6D[/latex], [latex]KH[/latex].

Which of [latex]1[/latex] or [latex]2[/latex] 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 {[latex]HH[/latex], [latex]HT[/latex], [latex]TH[/latex], [latex]TT[/latex]} where [latex]T[/latex] = tails and [latex]H[/latex] = heads. The outcomes are [latex]HH[/latex], [latex]HT[/latex], [latex]TH[/latex], and [latex]TT[/latex]. The outcomes HT and TH are different. The [latex]HT[/latex] means that the first coin showed heads and the second coin showed tails. The [latex]TH[/latex] means that the first coin showed tails and the second coin showed heads.

• Let [latex]A[/latex] = the event of getting at most one tail. (At most one tail means zero or one tail.) Then [latex]A[/latex] can be written as {[latex]HH[/latex], [latex]HT[/latex], [latex]TH[/latex]}. The outcome [latex]HH[/latex] shows zero tails.[latex]HT[/latex] and [latex]TH[/latex] each show one tail.
• Let [latex]B[/latex] = the event of getting all tails. [latex]B[/latex] can be written as {[latex]TT[/latex]}. [latex]B[/latex] is the complement of [latex]A[/latex], so [latex]B = A'[/latex]. Also, [latex]P(A) + P(B)[/latex] = [latex]P(A) + P(A') = 1[/latex].
• The probabilities for [latex]A[/latex] and for [latex]B[/latex] are [latex]P(A) = \displaystyle\frac{{3}}{{4}}[/latex] and [latex]P(B) = \displaystyle\frac{{1}}{{4}}[/latex].
• Let [latex]C[/latex] = the event of getting all heads. [latex]C[/latex] = {[latex]HH[/latex]}. Since [latex]B = {TT}[/latex], [latex]P(B \text{ AND } C) = 0[/latex]. [latex]B[/latex] and [latex]C[/latex] are mutually exclusive. ([latex]B[/latex] and [latex]C[/latex] have no members in common because you cannot have all tails and all heads at the same time.)
• Let [latex]D[/latex] = event of getting more than one tail. [latex]D = {TT}[/latex]. [latex]P(D) = \displaystyle\frac{{1}}{{4}}[/latex]
• Let [latex]E[/latex] = event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) [latex]E = {HT,HH}[/latex]. [latex]P(E) = \displaystyle\frac{{2}}{{4}}[/latex]
• Find the probability of getting at least one (one or two) tail in two flips. Let [latex]F[/latex] = event of getting at least one tail in two flips.[latex]F = {HT, TH, TT}[/latex]. [latex]P(F) = \displaystyle\frac{{3}}{{4}}[/latex]

Example

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

1. Let [latex]F[/latex] = the event of getting at most one tail (zero or one tail).
2. Let [latex]G[/latex] = the event of getting two faces that are the same.
3. Let [latex]H[/latex] = the event of getting a head on the first flip followed by a head or tail on the second flip.
4. Are [latex]F[/latex] and [latex]G[/latex] mutually exclusive?
5. Let [latex]J[/latex] = the event of getting all tails. Are [latex]J[/latex] and [latex]H[/latex] mutually exclusive?

Example

Roll one fair, six-sided die. The sample space is {[latex]1, 2, 3, 4, 5, 6[/latex]}. Let event
[latex]A[/latex] = a face is odd. Then [latex]A[/latex] = {[latex]1, 3, 5[/latex]}. Let event [latex]B[/latex] = a face is even. Then [latex]B[/latex] = {[latex]2, 4, 6[/latex]}.

• Find the complement of [latex]A[/latex], [latex]A'[/latex]. The complement of [latex]A[/latex], [latex]A'[/latex], is [latex]B[/latex] because [latex]A[/latex] and [latex]B[/latex] together make up the sample space. [latex]P(A) + P(B) = P(A) + P(A') = 1[/latex]. Also, [latex]P(A) = \displaystyle\frac{{3}}{{6}}[/latex] and [latex]P(B) = \displaystyle\frac{{3}}{{6}}[/latex].
• Let event [latex]C[/latex] = odd faces larger than two. Then [latex]C[/latex] = {[latex]3, 5[/latex]}. Let event [latex]D[/latex] = all even faces smaller than five. Then [latex]D[/latex] = {[latex]2, 4[/latex]}. [latex]P(C \text{ AND } D) = 0[/latex] because you cannot have an odd and even face at the same time. Therefore, [latex]C[/latex] and [latex]D[/latex] are mutually exclusive events.
• Let event [latex]E[/latex] = all faces less than five. [latex]E[/latex] = {[latex]1, 2, 3, 4[/latex]}.

Are [latex]C[/latex] and [latex]E[/latex] mutually exclusive events? (Answer yes or no.) Why or why not?

• Find [latex]P(C|A)[/latex]. This is a conditional probability. Recall that the event [latex]C[/latex] is {[latex]3, 5[/latex]} and event [latex]A[/latex] is {[latex]1, 3, 5[/latex]}. To find [latex]P(C|A)[/latex], find the probability of [latex]C[/latex] using the sample space [latex]A[/latex]. You have reduced the sample space from the original sample space {[latex]1, 2, 3, 4, 5, 6[/latex]} to {[latex]1, 3, 5[/latex]}. So, [latex]P(C|A) = \displaystyle\frac{{2}}{{3}}[/latex].

Example

Let event [latex]G[/latex] = taking a math class. Let event [latex]H[/latex] = taking a science class. Then, [latex]G[/latex] AND [latex]H[/latex] = taking a math class and a science class. Suppose [latex]P(G) = 0.6[/latex], [latex]P(H) = 0.5[/latex], and [latex]P(G \text{ AND } H) = 0.3[/latex]. Are [latex]G[/latex] and [latex]H[/latex] independent?

If [latex]G[/latex] and [latex]H[/latex] are independent, then you must show ONE of the following:

• [latex]P(G|H) = P(G)[/latex]
• [latex]P(H|G) = P(H)[/latex]
• [latex]P(G \text{ AND } H) = P(G)P(H)[/latex]

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 [latex]P(G|H) = P(G)[/latex].
2. Show [latex]P(G \text{ AND } H) = P(G)P(H)[/latex].

Since [latex]G[/latex] and [latex]H[/latex] 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 [latex]P(H|G) = P(H)[/latex] to show that [latex]G[/latex] and [latex]H[/latex] are independent events.

Example

Let event [latex]C[/latex] = taking an English class. Let event [latex]D[/latex] = taking a speech class.

Suppose [latex]P(C) = 0.75[/latex], [latex]P(D) = 0.3[/latex], [latex]P(C|D) = 0.75[/latex] and [latex]P(C \text{ AND } D) = 0.225[/latex].

Justify your answers to the following questions numerically.

1. Are [latex]C[/latex] and [latex]D[/latex] independent?
2. Are [latex]C[/latex] and [latex]D[/latex] mutually exclusive?
3. What is [latex]P(D|C)[/latex]?

Example

In a box there are three red cards and five blue cards. The red cards are marked with the numbers [latex]1, 2[/latex], and [latex]3[/latex], and the blue cards are marked with the numbers [latex]1, 2, 3, 4[/latex], and [latex]5[/latex]. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.

Let [latex]R[/latex] = red card is drawn, [latex]B[/latex] = blue card is drawn, [latex]E[/latex] = even-numbered card is drawn.

The sample space [latex]S = R1, R2, R3, B1, B2, B3, B4, B5[/latex]. [latex]S[/latex] has eight outcomes.

• [latex]P(R) = \displaystyle\frac{{3}}{{8}}[/latex].[latex]P(B) = \displaystyle\frac{{5}}{{8}}[/latex]. [latex]P(R \text{ AND } B) = 0[/latex]. (You cannot draw one card that is both red and blue.)
• [latex]P(E) = \displaystyle\frac{{3}}{{8}}[/latex]. (There are three even-numbered cards, [latex]R2, B2, \text{ and } B4[/latex].)
• [latex]P(E|B) = \displaystyle\frac{{2}}{{5}}[/latex]. (There are five blue cards: [latex]B1, B2, B3, B4, \text{ and } B5[/latex]. Out of the blue cards, there are two even cards; [latex]B2 \text{ and } B4[/latex].)
• [latex]P(B|E) = \displaystyle\frac{{2}}{{3}}[/latex]. (There are three even-numbered cards: [latex]R2, B2, \text{ and } B4[/latex]. Out of the even-numbered cards, to are blue; [latex]B2 \text{ and }B4[/latex].)
• The events [latex]R[/latex] and [latex]B[/latex] are mutually exclusive because [latex]P(R \text{ AND } B) = 0[/latex].
• Let [latex]G[/latex] = card with a number greater than [latex]3[/latex]. [latex]G[/latex] = {[latex]B4, B5[/latex]}. [latex]P(G) = \displaystyle\frac{{2}}{{8}}[/latex]. Let H = blue card numbered between one and four, inclusive. [latex]H[/latex] = {[latex]B1, B2, B3, B4[/latex]}. [latex]P(G|H) = \frac{1}{4}[/latex]. (The only card in [latex]H[/latex] that has a number greater than three is [latex]B4[/latex].) Since [latex]\frac{2}{8} = \frac{1}{4}[/latex], [latex]P(G) = P(G|H)[/latex], which means that [latex]G[/latex] and [latex]H[/latex] are independent.

Example

In a particular college class, [latex]60[/latex]% 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, [latex]75[/latex]% have long hair. Let [latex]F[/latex] be the event that a student is female. Let [latex]L[/latex] 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:
• [latex]P(F) = 0.60[/latex]; [latex]P(L) = 0.50[/latex]
• [latex]P(F \text{ AND } L) = 0.45[/latex]
• [latex]P(L|F) = 0.75[/latex]

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 [latex]P(F|L)[/latex] yet, so you cannot use the second condition.

Example

1. Toss one fair coin (the coin has two sides, [latex]H[/latex] and [latex]T[/latex]). The outcomes are ________. Count the outcomes. There are ____ outcomes.
2. Toss one fair, six-sided die (the die has [latex]1, 2, 3, 4, 5[/latex] or [latex]6[/latex] 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 [latex]H1[/latex] and [latex]T6[/latex].)
5. Event [latex]A[/latex] = heads ([latex]H[/latex]) on the coin followed by an even number ([latex]2, 4, 6[/latex]) on the die.
   [latex]A[/latex] = {_________________}. Find [latex]P(A)[/latex].
6. Event [latex]B[/latex] = heads on the coin followed by a three on the die. [latex]B[/latex] = {________}. Find [latex]P(B)[/latex].
7. Are [latex]A[/latex] and [latex]B[/latex] mutually exclusive? (Hint: What is [latex]P(A \text{ AND } B)[/latex]? If [latex]P(A \text{ AND } B) = 0[/latex], then [latex]A[/latex] and [latex]B[/latex] are mutually exclusive.)
8. Are [latex]A[/latex] and [latex]B[/latex] independent? (Hint: Is [latex]P(A \text{ AND } B) = P(A)P(B)[/latex]? If [latex]P(A \text{ AND } B) = P(A)P(B)[/latex], then [latex]A[/latex] and [latex]B[/latex] are independent. If not, then they are dependent).