Introduction to Statistics · Chapter 3 · Probability Topics

Probability Topics

A lab, not a reading: pick two candies at a time under two different rules, work out what each rule predicts, then run it yourself and see how close twelve trials come.

bookSHelf · Introduction to Statistics · §3.6 · a self-paced section

Probability Topics · bookSHelf Intro Stats§3.6

Learning objectives — by the end of this section you will be able to

Objectives

Compute a theoretical probability for a two-pick experiment, both with and without replacement Definitions 3.6.1–3.6.2
Run the experiment yourself and compute the matching empirical probability from your own record of results Tables 3.6.4–3.6.5
Compare the two estimates and explain, in terms of the number of trials, why they differ §3.6.9
Predict how the gap between an empirical estimate and its theoretical target behaves as the number of trials grows Try It Now 3.6.9

3.6

§3.6.1 — theoretical and empirical probability

Every probability in this lab has two faces: one you can compute from the bag before you ever touch it, and one you can only get by picking candies and keeping score.

Neither number is the "right" one — they answer different questions, and the whole lab is about watching them disagree.

Probability Topics · bookSHelf Intro Stats§3.6

The probability the bag already knows

Definition 3.6.1 — Theoretical Probability

The theoretical probability of an event is the ratio of the number of outcomes in that event to the total number of outcomes in the sample space, computed from the known makeup of the experiment:

P(\text{event}) = \frac{\text{number of outcomes in the event}}{\text{total number of outcomes}}

Definition 3.6.1: counted off the bag — 5 of the 40 candies are red, so P(red) = 5/40 without a single pick.

It requires no trials, because it comes from the structure of the experiment rather than from doing it — it is what the experiment should do in the long run.

Probability Topics · bookSHelf Intro Stats§3.6

The probability the trials produced

Definition 3.6.2 — Empirical Probability

The empirical probability of an event is the ratio of the number of trials in which the event actually occurred to the total number of trials performed:

P(\text{event}) = \frac{\text{number of trials in which the event occurred}}{\text{total number of trials}}

Definition 3.6.2: scored off the trials — 4 of 12 recorded picks were doubles, so P(doubles) = 4/12.

It requires no knowledge of the bag's contents — you get it by running the thing and keeping score. It is what the experiment did, and it is also called a relative frequency.

Probability Topics · bookSHelf Intro Stats§3.6

Insight Note — pricing a coin flip

Two roads to the same number

You can say heads is one-half because a coin has two sides, or you can flip it 12 times, get 7 heads, and say seven-twelfths. Neither person is lying. The first described the coin; the second described an afternoon.

Probability Topics · bookSHelf Intro Stats§3.6

Your turn — commit before the reveal

Try It Now 3.6.1

Try It Now 3.6.1 — Erin's die

Erin rolls a standard six-sided die 20 times. Four of her rolls come up 3. Write the theoretical probability of rolling a 3 and the empirical probability of rolling a 3 from this run, and say which one would change if she rolled another 20 times.

Step 1 — theoretical. A fair die has six equally likely faces and exactly one is a 3: $P(\text{roll a } 3) = \frac{1}{6}$ . This comes from the die's structure and does not change with more rolls.

Step 2 — empirical. Four of the 20 rolls came up 3: $P(\text{roll a } 3) = \frac{4}{20}$ .

Answer: theoretical $\frac{1}{6}$ , empirical $\frac{4}{20}$ — only the empirical value moves on a re-run; the theoretical value was never about any particular run.

Probability Topics · bookSHelf Intro Stats§3.6

§3.6.2 — building your own bag

Counting Your Population

Count out 40 mixed-color M&Ms — about one small bag's worth — and record how many of each color you have. These counts are your population: everything you compute on paper for the rest of the lab depends on them, so count twice.

Every worked calculation from here on uses one specific demonstration bag, so you can see each step done in full. Those numbers are not your numbers — read the demonstration to learn the move, then redo it with your own counts.

Probability Topics · bookSHelf Intro Stats§3.6

§3.6.2 — recording what you count

Table 3.6.1 — your population

Color	Quantity
Yellow (Y)
Green (G)
Blue (BL)
Brown (B)
Orange (O)
Red (R)

Table 3.6.1: Population — your own color counts.

Probability Topics · bookSHelf Intro Stats§3.6

§3.6.2 — the bag every worked example uses

Table 3.6.2 — the demonstration bag

Color	Quantity
Yellow (Y)	8
Green (G)	6
Blue (BL)	7
Brown (B)	5
Orange (O)	9
Red (R)	5
Total	40

Table 3.6.2: the demonstration bag used in every worked example below.

These numbers are not yours — read the demonstration to learn the move, then redo it with your own Table 3.6.1 counts.

Probability Topics · bookSHelf Intro Stats§3.6

Context Pause — reading the subscripts

The subscript is the pick number

Throughout this lab, $R_1$ means red on the first pick and $G_2$ means green on the second. The subscript is the pick number, not a second color. A double means both picks came out the same color.

$P(R_1 \text{ AND } G_2)$ and $P(G_1 \text{ AND } R_2)$ describe different events — say the event out loud in words before you write a fraction.

Probability Topics · bookSHelf Intro Stats§3.6

Your turn — read the bag

Try It Now 3.6.2

Try It Now 3.6.2 — not yellow

Using the demonstration bag in Table 3.6.2, how many M&Ms are not yellow, and what fraction of the bag is that?

Step 1 — total the bag. $8 + 6 + 7 + 5 + 9 + 5 = 40$ .

Step 2 — subtract the yellows. $40 - 8 = 32$ .

Answer: 32 of the 40 candies are not yellow, which is $\frac{32}{40}$ of the bag.

Probability Topics · bookSHelf Intro Stats§3.6

§3.6.3 — the first column

Theoretical Probabilities With Replacement

For each of the seven events in Table 3.6.3, compute the probability twice — once assuming the first M&M goes back before the second pick, once assuming it does not. This subsection fills the first column; leave every answer unreduced.

With replacement the cup is full again before the second pick, so both picks face the same 40 candies — $40 \cdot 40 = 1600$ ordered pairs, and the two picks are independent.

Probability Topics · bookSHelf Intro Stats§3.6

§3.6.3 — seven events, two designs

Table 3.6.3 — theoretical probabilities

Event	With Replacement	Without Replacement
$P(\text{2 reds})$
$P(R_1 B_2 \text{ OR } B_1 R_2)$
$P(R_1 \text{ AND } G_2)$
$P(G_2 \mid R_1)$
$P(\text{no yellows})$
$P(\text{doubles})$
$P(\text{no doubles})$

Table 3.6.3: compute these from your own Table 3.6.1 counts.

Probability Topics · bookSHelf Intro Stats§3.6

Your turn — one pair, with replacement

Try It Now 3.6.3

Try It Now 3.6.3 — brown then orange

Using the demonstration bag with replacement, compute $P(B_1 \text{ AND } O_2)$ — brown on the first pick, orange on the second. Leave the answer unreduced.

Step 1. The bag has 5 brown and 9 orange out of 40.

Step 2. The brown goes back in, so the second pick still faces all 40: $\frac{5}{40} \cdot \frac{9}{40} = \frac{45}{1600}$ .

Answer: $\frac{45}{1600}$ .

Probability Topics · bookSHelf Intro Stats§3.6

Worked example — the full with-replacement column

Example 3.6.1

Example 3.6.1 — filling in the with-replacement column

Tayen's demonstration bag has 8 yellow, 6 green, 7 blue, 5 brown, 9 orange, and 5 red, for a total of 40. Help them compute all seven theoretical probabilities with replacement.

Every row multiplies two fractions over 40, because the first candy always goes back. For doubles, add all six same-color products; no doubles is the complement over 1600.

$P(G_2 \mid R_1)$ only asks about the second pick — the red went back in, so the cup still holds all 6 greens out of 40, giving $\frac{6}{40}$ directly, with no second factor to multiply.

Event	With Rep.
2 reds	25/1600
$R_1B_2$ or $B_1R_2$	50/1600
$R_1$ and $G_2$	30/1600
$G_2 \mid R_1$	6/40
no yellows	1024/1600
doubles	280/1600
no doubles	1320/1600

Probability Topics · bookSHelf Intro Stats§3.6

§3.6.4 — the second column

Theoretical Probabilities Without Replacement

Redo the same seven events with the first candy left out of the cup. The first pick is unchanged, but every second pick now faces 39 candies instead of 40 — one short in whichever color left the cup.

That gives $40 \cdot 39 = 1560$ ordered pairs, and it makes the two picks dependent — you need the full multiplication rule, $P(A \text{ AND } B) = P(A)\,P(B \mid A)$ .

Probability Topics · bookSHelf Intro Stats§3.6

Your turn — the same pair, without replacement

Try It Now 3.6.4

Try It Now 3.6.4 — brown then orange, redone

Redo Try It Now 3.6.3 without replacement: compute $P(B_1 \text{ AND } O_2)$ , and say why the numerator did not change even though the design did.

Step 1. Brown first is $\frac{5}{40}$ . With a brown gone, the cup holds 39, and all 9 oranges are still there: $\frac{9}{39}$ .

Step 2. $\frac{5}{40} \cdot \frac{9}{39} = \frac{45}{1560}$ .

Answer: $\frac{45}{1560}$ — same numerator as with replacement, because removing a brown candy never touches the orange count; only the pool it's drawn from shrank.

Probability Topics · bookSHelf Intro Stats§3.6

Worked example — the full without-replacement column

Example 3.6.2

Example 3.6.2 — filling in the without-replacement column

Wes is redoing the same demonstration bag without replacement. Compute all seven theoretical probabilities the way he would.

Removing the first candy shrinks the second pick's pool to 39 — and shrinks that color's own count by one, if the second pick asks for the same color again.

For doubles, each color's second candy must be one of the remaining candies of that color: $8{\cdot}7 + 6{\cdot}5 + 7{\cdot}6 + 5{\cdot}4 + 9{\cdot}8 + 5{\cdot}4 = 240$ .

Event	Without Rep.
2 reds	20/1560
$R_1B_2$ or $B_1R_2$	50/1560
$R_1$ and $G_2$	30/1560
$G_2 \mid R_1$	6/39
no yellows	992/1560
doubles	240/1560
no doubles	1320/1560

Probability Topics · bookSHelf Intro Stats§3.6

The headline result — §3.6.5

Removing the first candy never touches a different-color pair — only same-color counts and the total move.

Both no-doubles rows read 1320: $\frac{1320}{1600}$ with replacement, $\frac{1320}{1560}$ without.

Banning replacement removes exactly the 40 ordered pairs where you'd have drawn the very same candy twice — and every one of those was a double. So the doubles count falls by 40 and the total falls by 40, leaving the two-different-colors count untouched.

† The general pattern: removing the first candy can only affect the second pick's count for that candy's own color. Ask for a different color second, and only the denominator moves; ask for the same color, and both move.

Probability Topics · bookSHelf Intro Stats§3.6

Insight Note — one shelf, not the whole cabinet

The rule change only touched one shelf

Banning replacement removes exactly the outcomes where you pick the same physical candy twice. Those were all doubles, so the non-doubles shelf is untouched — 1320 either way.

Probability Topics · bookSHelf Intro Stats§3.6

Your turn — predict without computing

Try It Now 3.6.5

Try It Now 3.6.5 — Alexis predicts first

Predict whether each is larger with replacement or without, for the demonstration bag: (a) $P(\text{2 reds})$ , and (b) $P(R_1 \text{ AND } G_2)$ .

(a) Same color twice. Removing the first red shrinks the numerator (25→20, a 20% drop) and the denominator (1600→1560, only 2.5%). The numerator loses more — without replacement is smaller.

(b) Two different colors. Removing a red takes nothing from the greens, so the numerator holds at 30 while the denominator drops — without replacement is larger.

Probability Topics · bookSHelf Intro Stats§3.6

§3.6.6 — the candy half

Running the Experiment

First pass — with replacement. Pick one M&M, record its color, put it back, then pick and record the second. Put that one back too. Do this 12 times.
Second pass — without replacement. Pick one M&M and record it — do not put it back — then pick and record the second. Return both before starting the next pair. Do this 12 times as well.

Returning both candies before the next pair is what makes the 12 pairs comparable to each other — each pair starts from a full cup of 40.

Probability Topics · bookSHelf Intro Stats§3.6

§3.6.6 — your record sheet

Table 3.6.4 — your empirical results

Pick	With Replacement	Without Replacement
1
2
⋮
12

Table 3.6.4: your 12 pairs under each design (60 total values).

Probability Topics · bookSHelf Intro Stats§3.6

Your turn — a bag with only one red

Try It Now 3.6.6

Try It Now 3.6.6 — can a pair be (R, R)?

Suppose you run the without-replacement pass with a bag that holds only one red candy. Could one of your pairs come out (R, R)? Explain what that tells you about reading a double.

Step 1. Without replacement, the two candies in a pair are always two different physical objects — the first is removed before the second is drawn.

Step 2. With only one red, once it's out there's no red left to draw — (R, R) is impossible.

Answer: no. A double means the two candies matched in color, not that the same candy was drawn twice — without replacement it always means two distinct candies sharing a color.

Probability Topics · bookSHelf Intro Stats§3.6

Worked example — a completed record sheet

Example 3.6.3

Example 3.6.3 — Malik's 24 pairs

Malik and his partner Devon ran both passes with the demonstration bag. His record is the input for every empirical calculation ahead.

With Replacement (picks 2, 6, 8, 12 are doubles — pick 6 drew the same red candy twice, which still counts)

1 (R,G) 2 (O,O) 3 (Y,B) 4 (BL,R) 5 (G,O) 6 (R,R) 7 (Y,O) 8 (BL,BL) 9 (O,G) 10 (B,Y) 11 (R,B) 12 (G,G)

Without Replacement (picks 2, 6, 10 are doubles — always two different candies of the same color)

1 (G,R) 2 (O,O) 3 (B,O) 4 (R,G) 5 (Y,G) 6 (R,R) 7 (BL,Y) 8 (G,O) 9 (R,B) 10 (Y,Y) 11 (B,G) 12 (O,BL)

Probability Topics · bookSHelf Intro Stats§3.6

§3.6.7 — counting from your record

Counting Your Empirical Probabilities

Go through your Table 3.6.4 record and count. For six of the seven events, count how many of your 12 pairs satisfied it and write that count over 12 — leave it unreduced.

The seventh event, $P(G_2 \mid R_1)$ , works differently and gets its own subsection next.

Probability Topics · bookSHelf Intro Stats§3.6

§3.6.7 — the counted version of Table 3.6.3

Table 3.6.5 — empirical probabilities

Event	With Replacement	Without Replacement
$P(\text{2 reds})$
$P(R_1 B_2 \text{ OR } B_1 R_2)$
$P(R_1 \text{ AND } G_2)$
$P(G_2 \mid R_1)$
$P(\text{no yellows})$
$P(\text{doubles})$
$P(\text{no doubles})$

Table 3.6.5: count these from your own Table 3.6.4 record.

Probability Topics · bookSHelf Intro Stats§3.6

Your turn — does the direction match?

Try It Now 3.6.7

Try It Now 3.6.7 — Rosalia checks Malik's doubles

Malik's empirical $P(\text{doubles})$ came out higher with replacement ( $\frac{4}{12}$ ) than without ( $\frac{3}{12}$ ). Does that match what the theoretical values predicted?

Step 1. Theoretical: $\frac{280}{1600} = 0.1750$ with replacement, $\frac{240}{1560} \approx 0.1538$ without — with-replacement is higher, and so is his data. The direction matches.

Answer: yes, but the theoretical gap (≈0.021) is smaller than one trial's worth ( $\frac{1}{12} \approx 0.083$ ) — 12 trials cannot reliably detect a difference that small, so this is encouraging, not confirming.

Probability Topics · bookSHelf Intro Stats§3.6

Worked example — counting Malik's record

Example 3.6.4

Example 3.6.4 — the six out-of-12 rows

Fill in the six out-of-12 rows of the empirical table from the 24 pairs in Example 3.6.3.

Event	With Rep.	Without Rep.
2 reds	1/12	1/12
$R_1B_2$ or $B_1R_2$	1/12	1/12
$R_1$ and $G_2$	1/12	1/12
no yellows	9/12	9/12
doubles	4/12	3/12
no doubles	8/12	9/12

Probability Topics · bookSHelf Intro Stats§3.6

Try it in rāSHio — let it do the tallying

Enter your results as a column and choose Graph → Frequency Table. It counts each distinct value and reports the matching relative frequency — the same pair of columns you're building by hand here. Count yourself first, then use the table to check your counts add back to the number of trials you ran.

Probability Topics · bookSHelf Intro Stats§3.6

§3.6.8 — the row that isn't out of 12

Reading a Conditional Probability From Your Record

The row $P(G_2 \mid R_1)$ is not out of 12 — this is the step most groups get wrong. A conditional probability only concerns the trials where the given condition actually happened.

How many of your 12 pairs started with red? That count is your denominator.
How many of those had green on the second pick? That count is your numerator.

Probability Topics · bookSHelf Intro Stats§3.6

Your turn — Jordan checks the conditional row

Try It Now 3.6.8

Try It Now 3.6.8 — both of Malik's passes

Compute the empirical $P(G_2 \mid R_1)$ for both of Malik's passes in Example 3.6.3, and explain the denominator you used.

With replacement. Pairs starting red: picks 1, 6, 11 — 3 pairs, 1 with green second: $P(G_2 \mid R_1) = \frac{1}{3}$ .

Without replacement. Pairs starting red: picks 4, 6, 9 — again 3 pairs, again 1 with green second: $\frac{1}{3}$ .

Answer: $\frac{1}{3}$ in both passes — the other nine pairs never satisfied $R_1$ , so including them would answer $P(R_1 \text{ AND } G_2)$ , a different question.

3.6

§3.6.9 — put the tables side by side

You now have two full tables of numbers for the same seven events. Put them side by side and look at where they agree and where they do not.

Probability Topics · bookSHelf Intro Stats§3.6

Your turn — what changes at 240 trials?

Try It Now 3.6.9

Try It Now 3.6.9 — Mai runs 240 instead of 12

What values could her empirical $P(\text{doubles})$ take, and would you expect it to land closer to the theoretical 0.1750?

Step 1. With 240 trials the step size drops to about 0.0042 (from 0.0833) — any multiple of $\frac{1}{240}$ .

Step 2. $0.1750 \times 240 = 42$ , a whole number — $\frac{42}{240}$ hits the target exactly, something 12 trials never could.

Answer: any multiple of $\frac{1}{240}$ ; yes — the Law of Large Numbers pulls long-run relative frequency toward the theoretical value.

Probability Topics · bookSHelf Intro Stats§3.6

Worked example — theoretical against empirical

Example 3.6.5

Example 3.6.5 — two gaps, one resolution

For Malik's bag and record, compare the with-replacement estimates of $P(\text{no yellows})$ and $P(\text{doubles})$ in decimal form.

Event	Theoretical	Empirical	Gap
no yellows	0.6400	0.7500	+0.1100
doubles	0.1750	0.3333	+0.1583

One extra matching trial out of 12 moves the empirical estimate by $\frac{1}{12} \approx 0.0833$ — both gaps here are within about two of those steps. With only 12 trials the empirical value can land on just 13 possible numbers, so it physically cannot sit closer to the truth, however carefully the experiment was run.

Probability Topics · bookSHelf Intro Stats§3.6

Context Pause — why polls quote a sample size

The resolution is set by how many trials paid for it

A poll of 12 people cannot report 43% support, because 43% is not a multiple of one-twelfth. Run 240 trials instead of 12 and the step size drops to $\frac{1}{240}$ , which lets the empirical value land near the theoretical one — and the Law of Large Numbers says it will.

Probability Topics · bookSHelf Intro Stats§3.6

Try it in rāSHio — simulate what you can't hand-draw

You are not going to draw 240 pairs of M&Ms by hand, but you can simulate them. Open rāSHio and choose File → Random Numbers…, set the range 1 through 40, leave No repeats UNCHECKED so a color can repeat, and ask for far more than a handful. Compare a short column against a long one — the long one is not more random, it is just further along.

Probability Topics · bookSHelf Intro Stats§3.6

Figure 3.6.1

The rāSHio random-number dialog

File → Random Numbers…

Set a minimum, a maximum, and how many numbers to draw, then press Generate. Widen the range and the count to simulate a long run.

Figure 3.6.1: set a range and a count, then Generate. Widen the range and the count to simulate a long run.

Probability Topics · bookSHelf Intro Stats§3.6

§3.6.10 — discuss with your group

Discussion Questions

Answer these with your group, using your own two tables. Every one is a judgement call — what matters is whether you can point at your own numbers to defend the answer.

Why are the With-Replacement and Without-Replacement theoretical probabilities different? Name the specific step where the two designs stop agreeing, and say what physically causes it.
Convert $P(\text{no yellows})$ to decimal for both the theoretical and empirical With-Replacement values (round to four places) — are they closer or farther apart than you expected?
If you increased the picks from 12 to 240, why would the empirical values change? Be specific about which part of the fraction moves and which does not.
Would that change make the empirical and theoretical probabilities closer together — and is that a guarantee or a strong expectation?
Explain the difference between what $P(G_1 \text{ AND } R_2)$ and $P(R_1 \mid G_2)$ represent: how many trials does each get computed over, and why is it not the same number?

Probability Topics · bookSHelf Intro Stats§3.6

Key Terms — this lab's vocabulary

Key Terms

theoretical probability — a probability computed from the known structure of an experiment, the ratio of favorable outcomes to total outcomes, without running any trials.

empirical probability — a probability estimated from data: the ratio of trials in which an event occurred to the total trials; also called a relative frequency.

trial — one full repetition of the experiment; in this lab, one pick of two M&Ms.

double — an outcome in which both picks in a trial come out the same color.

sampling with replacement — the first item is returned before the second is drawn, so both picks face the same population.

sampling without replacement — the first item is not returned before the second is drawn, so the second pick faces a population one item smaller.

3.6

Probability Topics · bookSHelf Intro Stats§3.6

§3.6 — conclusions

What §3.6 leaves you with

The core idea

Theoretical probability comes from the bag's known makeup; empirical probability comes from running the experiment and keeping score. They almost never agree exactly — the disagreement is information about how many trials you ran, not a mistake.

Watch for

A 12-trial estimate has a step size of $\frac{1}{12}$ , so it cannot sit closer to the truth than about half a step, however carefully you run it. More trials shrink the step and let the Law of Large Numbers pull the empirical value toward the theoretical one.

Next: Chapter 4 — Random Variables, starting with the probability distribution function for a discrete random variable.