Experimental Probability

Section 1 of 11

Relative frequency

Relative frequency is the experimental probability — what we actually observe after running an experiment.

Must learn

defRelative frequency $= \dfrac{\text{number of times the event happened}}{\text{total number of trials}}$

Worked example — biased coin?

A coin is tossed 100 times. It comes up 56 tails and 44 heads. Find the relative frequency.

$P(T) = \dfrac{56}{100} = 0.56$

$P(H) = \dfrac{44}{100} = 0.44$

Is the coin biased?

First instinct: "Yes — $P(T)$ should be $0.5$, and $0.56$ is much bigger."

Better answer: Maybe — not sure. Not enough trials (experiments).

With only 100 tosses, 56T/44H could easily happen by chance with a fair coin. You'd want thousands of trials before declaring bias.

YOU TRY · 1

A die is rolled 200 times. A "6" comes up 38 times. Find the relative frequency of rolling a 6.

Number of 6s over total rolls.

$P(6) = \dfrac{38}{200}$

$= 0.19$

$0.19$

YOU TRY · 2

In a survey of 250 people, 175 said they preferred tea over coffee. What is the relative frequency of preferring tea?

Express as a decimal.

$P(\text{tea}) = \dfrac{175}{250}$

$= 0.7$

$0.7$

Section 2 of 11

Probability tables — finding the missing value

A biased die has this probability table:

Result	1	2	3	4	5	6
Prob	0.1	0.3	0.1	0.2	$x$	0.1

(i) Find $x$

All probabilities must sum to $1$.

$0.1 + 0.3 + 0.1 + 0.2 + x + 0.1 = 1$

$0.8 + x = 1$

$x = 0.2$

(ii) If the die is thrown 1000 times, how many 5s would you expect?

No. of 5s $= 0.2 \times 1000$

$= 200$

Must learn

1.$\sum P(x) = 1$ (all probabilities add to 1)

2.Expected count $= P(\text{event}) \times \text{number of trials}$

YOU TRY · 3

A spinner has 4 colours: red, blue, green, yellow. $P(\text{red}) = 0.3$, $P(\text{blue}) = 0.25$, $P(\text{green}) = 0.15$. Find $P(\text{yellow})$.

They must sum to 1.

$0.3 + 0.25 + 0.15 + P(\text{yellow}) = 1$

$0.7 + P(\text{yellow}) = 1$

$P(\text{yellow}) = 0.3$

$0.3$

YOU TRY · 4

Using the spinner from Q3, if you spin 500 times, how many reds would you expect?

Probability × number of trials.

No. of reds $= 0.3 \times 500$

$= 150$

$150$

YOU TRY · 5

A bag contains marbles of 3 colours. $P(\text{red}) = 0.4$, $P(\text{blue}) = y$, $P(\text{green}) = 2y$. Find $y$.

Sum to 1, then solve for $y$.

$0.4 + y + 2y = 1$

$0.4 + 3y = 1$

$3y = 0.6$

$y = 0.2$

Section 3 of 11

Expected Value — the idea

Toss a €2 coin. If it lands tails, you keep the coin (win €2). If it lands heads, it's lost (win €0).

What would you expect to win on average?

Expected value = the average you'd win per try, if you played many, many times.

Average $= \dfrac{0 + 2}{2}$

$=$ €$1$

So if you played this game many times, you'd average €1 per go. You won't win €1 on any one go — you'll win either €0 or €2 — but €1 is the long-run average.

Section 4 of 11

Expected Value — the dartboard

A dartboard is split into 4 equal quadrants, each worth a different amount. It costs €10 to play.

Each quadrant is equally likely, so $P = \tfrac{1}{4}$ for each.

Will I play? Work out the expected winnings first.

$E = \dfrac{2 + 5 + 8 + 20}{4}$

$= \dfrac{35}{4}$

$=$ €$8.75$

Or written as $\sum x\,P(x)$:

$E = \tfrac{1}{4}(2) + \tfrac{1}{4}(5) + \tfrac{1}{4}(8) + \tfrac{1}{4}(20)$

$=$ €$8.75$

Cost to play is €10. Average winnings are €8.75.

Loss per go $= 8.75 - 10 =$ −€$1.25$

On average you lose €1.25 every game. Don't play.

YOU TRY · 6

A board has 4 equal sections worth €1, €3, €6, €10. It costs €6 to play. What is the expected winnings? Should you play?

Average the four values. Compare to €6.

$E = \dfrac{1 + 3 + 6 + 10}{4} = \dfrac{20}{4}$

$=$ €$5$

Loss $= 5 - 6 =$ −€$1$. Don't play.

$E =$ €$5$, loss of €1 per go. Don't play.

Section 5 of 11

Expected Value — three sections

Now a board split into 3 equal sections: €15, €30, €5. Costs €20 to play.

3 equal sections, so $P = \tfrac{1}{3}$ each.

$E = \dfrac{15 + 30 + 5}{3} = \dfrac{50}{3}$

$=$ €$16.67$

Or using $\sum x\,P(x)$:

$E = \tfrac{1}{3}(15) + \tfrac{1}{3}(30) + \tfrac{1}{3}(5) = \dfrac{50}{3}$

Cost €20, average winnings €16.67.

Loss $= 20 - 16.67 =$ €$3.33$ per go

Don't play this either.

Section 6 of 11

The Expected Value formula

Now we put names on what we've been doing.

$x$ = event

$P(x)$ = probability of event

$E(x)$ = expected value

For events $x_1, x_2, \ldots, x_n$:

$E(x) = x_1 \cdot P(x_1) + x_2 \cdot P(x_2) + \cdots + x_n \cdot P(x_n)$

In short — the sum of $x \cdot P(x)$:

Must learn

1.$E(x) = \sum x \cdot P(x)$

2.$\sum P(x) = 1$

YOU TRY · 7

A game gives you €5 with probability 0.3, €10 with probability 0.5, and €20 with probability 0.2. Find $E(x)$.

$\sum x\,P(x)$ — multiply each event by its probability and add.

$E(x) = 5(0.3) + 10(0.5) + 20(0.2)$

$= 1.5 + 5 + 4$

$=$ €$10.50$

€$10.50$

YOU TRY · 8

Find $E(x)$ for: $x = 1, 2, 3, 4$ with probabilities $0.1, 0.2, 0.4, 0.3$.

Check probabilities sum to 1 first.

Check: $0.1 + 0.2 + 0.4 + 0.3 = 1$ ✓

$E(x) = 1(0.1) + 2(0.2) + 3(0.4) + 4(0.3)$

$= 0.1 + 0.4 + 1.2 + 1.2$

$= 2.9$

$2.9$

YOU TRY · 9

A raffle has tickets: €0 (prob 0.7), €5 (prob 0.2), €50 (prob 0.1). Find $E(x)$. If a ticket costs €8, is it worth buying?

Find $E(x)$ first, then compare to €8.

$E(x) = 0(0.7) + 5(0.2) + 50(0.1)$

$= 0 + 1 + 5$

$=$ €$6$

Loss $= 8 - 6 =$ €$2$ per ticket. Don't buy.

$E =$ €$6$. Don't buy — €2 loss per go.

Section 7 of 11

Solving for unknowns — using $E(x)$

Given the table:

$x$	1	2	3	4
$P(x)$	0.2	$p$	$q$	0.3

Given $E(x) = 2.5$, find $p$ and $q$.

Equation 1: probabilities sum to 1

$0.2 + p + q + 0.3 = 1$

$p + q = 0.5$

Equation 2: $E(x) = \sum x\,P(x) = 2.5$

$1(0.2) + 2p + 3q + 4(0.3) = 2.5$

$0.2 + 2p + 3q + 1.2 = 2.5$

$2p + 3q = 1.1$

Solve simultaneously

$p + q = 0.5 \quad \Rightarrow \quad 2p + 2q = 1$ (× 2)

Subtract from $2p + 3q = 1.1$:

$q = 0.1$

Then $p = 0.5 - 0.1$:

$p = 0.4, \quad q = 0.1$

YOU TRY · 10

Given $x = 0, 1, 2, 3$ with $P(x) = 0.1, a, b, 0.2$ and $E(x) = 1.5$, find $a$ and $b$.

Two equations: $\sum P = 1$ and $\sum xP = 1.5$.

Eq1: $0.1 + a + b + 0.2 = 1 \Rightarrow a + b = 0.7$

Eq2: $0(0.1) + 1a + 2b + 3(0.2) = 1.5$

$\Rightarrow a + 2b + 0.6 = 1.5 \Rightarrow a + 2b = 0.9$

Subtract: $b = 0.2$, so $a = 0.5$.

$a = 0.5, \quad b = 0.2$

YOU TRY · 11

For $x = 1, 2, 3$ with $P(x) = p, 0.4, q$ and $E(x) = 2.2$, find $p$ and $q$.

Same approach — two simultaneous equations.

Eq1: $p + 0.4 + q = 1 \Rightarrow p + q = 0.6$

Eq2: $1p + 2(0.4) + 3q = 2.2 \Rightarrow p + 3q = 1.4$

Subtract Eq1: $2q = 0.8 \Rightarrow q = 0.4$

So $p = 0.6 - 0.4 = 0.2$

$p = 0.2, \quad q = 0.4$

Section 8 of 11

Sigma Notation

$\Sigma$ (sigma) means "sum of".

How to read it

$\displaystyle\sum_{n=3}^{5} n^2$

The bottom tells you the start ($n = 3$).

The top tells you the end ($n = 5$).

Substitute each value into the expression and add.

$\displaystyle\sum_{n=3}^{5} n^2 = 3^2 + 4^2 + 5^2$

$= 9 + 16 + 25$

$= 50$

YOU TRY · 12

Evaluate $\displaystyle\sum_{n=1}^{4} n^2$.

Start at 1, end at 4, square each, add.

$1^2 + 2^2 + 3^2 + 4^2$

$= 1 + 4 + 9 + 16$

$= 30$

$30$

YOU TRY · 13

Evaluate $\displaystyle\sum_{n=2}^{5} 3n$.

Sub in $n = 2, 3, 4, 5$.

$3(2) + 3(3) + 3(4) + 3(5)$

$= 6 + 9 + 12 + 15$

$= 42$

$42$

Section 9 of 11

Big sums — the pairing trick

Evaluate $\displaystyle\sum_{n=1}^{100} n$ (with $n \in \mathbb{N}$).

$= 1 + 2 + 3 + 4 + \cdots + 98 + 99 + 100$

Pair the first with the last, the second with the second-last, and so on:

$(1 + 100) = 101$

$(2 + 99) = 101$

$(3 + 98) = 101$

$\vdots$

100 numbers means 50 pairs, each adding to 101.

$= 50 \times 101$

$= 5050$

The formula

This generalises into the arithmetic series formula:

$S_n = \dfrac{n}{2}\{2a + (n-1)d\}$

You'll meet this properly in Arithmetic Series.

Section 10 of 11

Sigma — more practice

Evaluate $\displaystyle\sum_{n=5}^{8} (2n + 1)$.

$= (2(5)+1) + (2(6)+1) + (2(7)+1) + (2(8)+1)$

$= 11 + 13 + 15 + 17$

$= 56$

YOU TRY · 14

Evaluate $\displaystyle\sum_{n=1}^{4} (3n - 2)$.

Sub in 1, 2, 3, 4.

$(3(1)-2) + (3(2)-2) + (3(3)-2) + (3(4)-2)$

$= 1 + 4 + 7 + 10$

$= 22$

$22$

YOU TRY · 15

Evaluate $\displaystyle\sum_{n=2}^{6} n(n+1)$.

Five terms — be careful with the multiplication.

$2(3) + 3(4) + 4(5) + 5(6) + 6(7)$

$= 6 + 12 + 20 + 30 + 42$

$= 110$

$110$

YOU TRY · 16

Use the pairing trick to find $\displaystyle\sum_{n=1}^{50} n$.

First + last gives the pair value. How many pairs?

$1 + 50 = 51$

50 numbers → 25 pairs

$= 25 \times 51$

$= 1275$

$1275$

SUM

The lot in one box

Experimental Probability toolkit

1.Relative frequency $= \dfrac{\text{hits}}{\text{trials}}$

2.$\sum P(x) = 1$ (probabilities sum to 1)

3.Expected count $= P \times \text{trials}$

4.$E(x) = \sum x \cdot P(x)$

5.Two unknowns? Use $\sum P = 1$ and $E(x)$ as simultaneous equations.

6.$\Sigma$ = sum of. Bottom = start, top = end.

7.$\displaystyle\sum_{n=1}^{100} n = 5050$ (pairing trick)

End of lesson

Experimental Probability — HL · Mathslive.ie