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, 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.TRY IT

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.

1. Suppose you know that the picked cards are Q of spades, K of hearts and Q of spades. Can you decide if the sampling was with or without replacement?
2. 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?

Solution:

1. With replacement
2. No

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, 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.

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?

Solution:

1. Without replacement
2. With 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) = [latex]\displaystyle\frac{{3}}{{4}}[/latex].
• 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) = [latex]\displaystyle\frac{{1}}{{4}}[/latex]
• 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) = [latex]\displaystyle\frac{{2}}{{4}}[/latex]
• 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) = [latex]\displaystyle\frac{{3}}{{4}}[/latex]

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?

Solution:

Look at the sample space in Example 3.

1. Zero (0) or one (1) tails occur when the outcomes HH, TH, HT show up. P(F) = [latex]\displaystyle\frac{{3}}{{4}}[/latex]
2. Two faces are the same if HH or TT show up. P(G) = [latex]\displaystyle\frac{{2}}{{4}}[/latex]
3. 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) = [latex]\displaystyle\frac{{2}}{{4}}[/latex]
4. F and G share HH so P(F AND G) is not equal to zero (0). F and G are not mutually exclusive.
5. 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.

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) = [latex]\displaystyle\frac{{3}}{{6}}[/latex].
• 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(CAND 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?

Solution:

No.
C = {3, 5} and E = {1, 2, 3, 4}. P(C AND E) = [latex]\displaystyle\frac{{1}}{{6}}[/latex]. To be mutually exclusive, 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) = [latex]\displaystyle\frac{{2}}{{3}}[/latex].

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 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)

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(GAND H) = P(G)P(H).

Solution:

1. One
2. Two

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.

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 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)?

Solution:

1. Yes, because P(C|D) = P(C).
2. No, because P(C AND D) is not equal to zero.
3. [latex]\displaystyle{P}{({D}{mid}{C})}=\frac{{{P}{({C} \text{ AND }{D})}}}{{{P}{({C})}}}=\frac{{0.225}}{{0.75}}={0.3}[/latex]

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) = [latex]\displaystyle\frac{{3}}{{8}}[/latex]. P(R AND B) = 0. (You cannot draw one card that is both red and blue.)
• P(E) = [latex]\displaystyle\frac{{3}}{{8}}[/latex]. (There are three even-numbered cards, R2, B2, and B4.)
• P(E|B) = [latex]\displaystyle\frac{{2}}{{5}}[/latex]. (There are five blue cards: B1, B2, B3, B4, and B5. Out of the blue cards, there are two even cards; B2 andB4.)
• P(B|E) = [latex]\displaystyle\frac{{2}}{{3}}[/latex]. (There are three even-numbered cards: R2, B2, and B4. Out of the even-numbered cards, to are blue; B2 andB4.)
• 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) = [latex]\displaystyle\frac{{2}}{{8}}[/latex], 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 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.

Solution 1:

Check whether P(F AND L) = P(F)P(L). We are given that P(F AND 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 AND L) does not equal P(F)P(L).

Solution 2:

Check whether P(L|F) equals P(L). We are given that P(L|F) = 0.75, but P(L) = 0.50; they are not equal. The events of being female and having long hair are not independent.

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

Solution:

1. H and T; 2
2. 1, 2, 3, 4, 5, 6; 6
3. 2(6) = 12
4. T1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6
5. A = {H2, H4, H6}; P(A) = [latex]\displaystyle\frac{{3}}{{12}}[/latex]
6. B = {H3}; P(B) = [latex]\displaystyle\frac{{1}}{{12}}[/latex]
7. Yes, because P(A AND B) = 0
8. P(A AND B) = 0.P(A)P(B) = [latex]\displaystyle{(\frac{{3}}{{12}})}{(\frac{{1}}{{12}})}[/latex]. P(A AND B) does not equal P(A)P(B), so A and B are dependent.