Consider an experiment where we draw two cards from a well-shuffled standard deck of cards without replacement. What’s the probability that the first card is an ace, and the second card is a 2?

Since we care about what the first card is, and what the second card is, the order in which we draw cards matters. Furthermore, because there are 4 suits that a card can belong to, each ace is considered distinct from the other three aces. The same applies to the twos.

The sample space consists of all pairs of cards where no card is repeated because we draw without replacement. Thus, we have that

|S| = 52 * 51.

The event we care about is the event

E = { (c₁, c₂) : c₁ is an ace ∧ c₂ is a two }.

Since there are 4 aces and 4 twos, we have that

|E| = 4 * 4.

Thus, the desired probability is

Consider an experiment where we draw two cards from a well-shuffled standard deck of cards without replacement. What’s the probability that one of the cards drawn is an ace, and the other is a 2?

Here, the ace can be first and the 2 can be second, or the ace can be second while the 2 is first. Both of these occurrences are mutually exclusive, so we have that

|E| = 4*4 + 4*4.

We have the same sample space from Example 1.6.1, meaning the probability for this event is

Consider an experiment where we are dealt a five-card hand from a well-shuffled standard deck of cards. What is the probability that we are dealt a full house; that is, what is the probability that our hand has three cards of one face value, and two cards of a different face value?

When it comes to hands, order is irrelevant. This means that our sample space for this experiment is all possible five-card hands, which is counted by the following.

We assume that no five-card hand is more likely to be dealt than any other five-card hand, and so the assumption of uniform randomness applies.

To calculate |E|, we can think in stages. First, we choose which of the 13 face values we want to represent the triple. There are 4 cards of the same face value, and we pick 3 of them without regard to order. Since there are 4 cards with the same face value, we can do this process in 13 different ways. Thus, there are

ways to choose the triple.

For the second stage, we can exclude the previous face value, leaving us with 12 face values left from which to choose. For each one of these remaining face values, we pick 2 cards from the 4 face values without regard to order, and so for the second stage, we have

choices. Now we have a full house hand. Since we broke this down into stages, the rule of product tells us that

Thus, the probability of getting a full house is

Consider an experiment where you are dealt a five card hand. What is the probability that you are dealt a straight?

Because order is irrelevant, our sample space consists of all 5 card hands where order is not relevant. As normal, we assume that all hands are equally likely, so we have that

To calculate |E|, notice that there are 10 kinds of straights:

5-high, 6-high, 7-high, … king-high, ace-high.

Notice that since each card in a straight has different face value, there are 4 choices of suit for each card in the hand. As such, we have that

|E| = 10 * 4⁵.

From this, we calculate the probability:

Consider an experiment where you shuffle a deck of cards and place it face down on the table. What is the probability that the ace of spades is closer to the top of the deck than the 2 of spades?

Notice that since we want to know the position of one card in relation to another card, order matters.

The sample space consists of all possible sequences of cards. Under the assumption of uniform randomness, we hae that

|S| = 52!.

We can calculate |E| by considering mutually exclusive cases. Suppose we number the cards where the top card is #1, and the bottom card is #52.

(♠, A) is #1 ⟹ (♠, 2) can in be in 1 of 51 spots
(♠, A) is #2 ⟹ (♠, 2) can in be in 1 of 50 spots
…
(♠, A) is #51 ⟹ (♠, 2) can in be in only be in position #52.

So, there are 51 possible locations for (♠, A), each yielding a smaller number of allowable positions for (♠, 2). However, notice that this only counts the two cards of interest. For each of these cases, there are 50! ways to arrange the other 50 cards. As such, we have that

We can now calculate the probability:

Suppose an urn contains 15 blue balls and 20 orange balls where all of the blue balls are all identical, and all orange balls are identical. Consider an experiment where you randomly draw 10 balls out of the urn without replacement. What’s the probability that you draw 5 balls of each color?

The sample space S consists of all ways to draw 10 balls out of the 15 + 20 = 35 balls from the urn. Order is not relevant here, so we have that

Because order is irrelevant, we can arbitrarily decide to count the number of ways to first select 5 blue balls from the 15 blue balls present, and then count the number of ways to select 5 orange balls from the 20 orange balls already present. The by the rule of product, we get that

And so the probability is

In Example 1.6.6, we regarded our selections as unordered, but we could also solve this problem using ordered selections.

In order to enumerate the sample space, note that there are eleven mutually exclusive cases to consider:

Blue Balls	Orange Balls
0	10
1	9
2	8
3	7
4	6
5	5
6	4
7	3
8	2
9	1
10	0

For each of these cases, let n_b represent the number of blue balls selected, and let n_o represent the number of orange balls selected. First, suppose we draw all n_b blue balls first, and then draw all n_o orange balls, so our selection looks like the following:

⬤,⬤ , …,⬤ , ⬤, ⬤, …, ⬤.

There are

ways to draw n_b blue balls and n_o orange balls where all blue balls come first.

For example, if n_b = 3, then (since we draw 10 balls,) n_o =7, and our drawing would look like this:

⬤, ⬤, ⬤, ⬤, ⬤, ⬤, ⬤, ⬤, ⬤, ⬤.

And there would be

ways to select those 3 blue balls and 7 orange balls.

Now, note that the blue balls don’t have to be first; they could be dispersed around among the orange balls. For n_b blue balls and n_o orange balls, there are

ways to arrange the drawing. So for the case where n_b = 3 and n_o =7, there are

such arrangements.

Finally, when we account for the fact that there are eleven cases to consider, we get that our sample space has

elements.

To calculate |E|, notice that we already know how to count arrangements with 5 blue balls and 5 orange balls, since that was one of the cases used to count |S|. As such, we have that

Finally, our probability is

This is the same answer we got previously!

Consider an experiment where we roll a die twice. What’s the probability that at least one roll is a six?

Our sample space S is going to consist of all ordered pairs from the Cartesian Product

S = {1, 2, 3, 4, 5, 6} ✕ {1, 2, 3, 4, 5, 6},

and so |S| = 36.

This is easy enough to list in a table:

Roll 1 \ Roll 2	1	2	3	4	5	6
1	(1, 1)	(1, 2)	(1, 3)	(1, 4)	(1, 5)	(1, 6)
2	(2, 1)	(2, 2)	(2, 3)	(2, 4)	(2, 5)	(2, 6)
3	(3, 1)	(3, 2)	(3, 3)	(3, 4)	(3, 5)	(3, 6)
4	(4, 1)	(4, 2)	(4, 3)	(4, 4)	(4, 5)	(4, 6)
5	(5, 1)	(5, 2)	(5, 3)	(5, 4)	(5, 5)	(5, 6)
6	(6, 1)	(6, 2)	(6, 3)	(6, 4)	(6, 5)	(6, 6)

The event E consists of all ordered pairs with at least one 6. We can color each of those cells green.

Roll 1 \ Roll 2	1	2	3	4	5	6
1	(1, 1)	(1, 2)	(1, 3)	(1, 4)	(1, 5)	(1, 6)
2	(2, 1)	(2, 2)	(2, 3)	(2, 4)	(2, 5)	(2, 6)
3	(3, 1)	(3, 2)	(3, 3)	(3, 4)	(3, 5)	(3, 6)
4	(4, 1)	(4, 2)	(4, 3)	(4, 4)	(4, 5)	(4, 6)
5	(5, 1)	(5, 2)	(5, 3)	(5, 4)	(5, 5)	(5, 6)
6	(6, 1)	(6, 2)	(6, 3)	(6, 4)	(6, 5)	(6, 6)

Here, we se that |E| = 11. Now we calculate the probability:

Consider an experiment where you roll a die four times. What’s the probability that at least one of the rolls is a six?

Similarly to Example 1.6.8, we have that

S = {1, 2, 3, 4, 5, 6} ✕ {1, 2, 3, 4, 5, 6} ✕ {1, 2, 3, 4, 5, 6} ✕ {1, 2, 3, 4, 5, 6}.

This means that |S| = 6⁴ = 1296, which is way too many elements to conveniently list. We know that E is the event where at least one roll is a 6. However, that means that E^C is the event that no roll is a 6. This means that

E^C = {1, 2, 3, 4, 5} ✕ {1, 2, 3, 4, 5} ✕ {1, 2, 3, 4, 5} ✕ {1, 2, 3, 4, 5}.

As such,

|E^C| = 5⁴ = 625.

Furthermore, because E ∪ E^C = S and E ∩ E^C = ∅, we have that

|E| = |S| – |E^C| = 1296 – 625 = 671.

Now we calculate the probability:

The senior class at a local high school is required to have completed a course in

A:	American History
B:	Biology
C:	Calculus

at some point during their high school education. Let x_s represent the number of students enrolled in some combination of courses during the final year; for example,

xB	= The total number of students enrolled in Biology
xAC	= The total number of students enrolled in American History and Calculus
xABC	= The total number of students enrolled in all three courses

Suppose we know the following information:

xA	= 90
xB	= 78
xC	= 105
xAB	= 34
xAC	= 35
xBC	= 36
xABC	= 10

Furthermore, we also know that this senior class has 200 students. From this information, we construct the following Venn Diagram:

Consider an experiment where we pick two senior students at random. What’s the probability they are both only enrolled in the American History course?

The sample space S consists of every 2-student pair selected from the 200 students available. The order in which we pick the students is irrelevant, so we have that

Next, the event E consists of essentially picking two students out of the 31 that are only enrolled in the American History course, and so we have that

And so, we finally have that