Probability and expected outcomes AI HL

IB Mathematics: Applications and Interpretation HL · Unit 4

Lesson 4.5 · Probability and expected outcomes

Describe possible outcomes, calculate probabilities and estimate how often an event may occur.

Hope of being with Jesus

As Jesus was dying on the cross, one of the criminals beside him asked to be remembered. Jesus answered: “Truly I tell you, today you will be with me in paradise.” Luke 23:43 (NIV)

This moment gives Christians a picture of hope: the promise of life with Jesus does not depend on human achievement, but on his mercy. We look forward to eternal life with God and trust the promise Jesus gave of being with him.

Learning objectives

By the end of this lesson, you should be able to:

identify trials, outcomes, events and sample spaces;
calculate probabilities when outcomes are equally likely;
estimate probability from relative frequency;
find the probability of a complementary event; and
calculate the expected number of occurrences of an event.

1. Trials, outcomes and events

A trial is one performance of a chance experiment, such as tossing a coin or rolling a die. An outcome is one possible result. An event is a specified outcome or collection of outcomes.

The sample space, often written $U$ , is the set of all possible outcomes of a trial or experiment.

Worked example · Tossing a coin three times

Let H represent heads and T represent tails. The sample space is:

$U=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$

There are 8 possible outcomes. The event “exactly two heads” is $\{HHT,HTH,THH\}$ , so it contains 3 outcomes.

$P(\text{exactly two heads})=\frac{3}{8}$

Organise outcomes carefully. A tree diagram, table or list can help you make sure every possible outcome is included once.

2. Theoretical probability

When all outcomes are equally likely, the probability of an event $A$ is the number of outcomes in $A$ divided by the total number of outcomes in the sample space.

$\boxed{P(A)=\frac{n(A)}{n(U)}}$

Probability is always between 0 and 1:

$P(A)=0$ means the event is impossible;
$P(A)=1$ means the event is certain; and
values between 0 and 1 describe different likelihoods.

Worked example · Rolling a fair die

A fair die has six equally likely outcomes: $\{1,2,3,4,5,6\}$ . The event “roll an even number” is $\{2,4,6\}$ .

$P(\text{even})=\frac{3}{6}=\boxed{\frac{1}{2}}$

The formula $P(A)=\dfrac{n(A)}{n(U)}$ applies when outcomes are equally likely. It should not be used automatically for a biased die, an unfair game or outcomes with unequal chances.

3. Experimental probability and relative frequency

For an experiment, the relative frequency of an event is the number of times it occurs divided by the total number of trials. It can be used to estimate probability.

$\boxed{\text{Relative frequency of }A= \frac{\text{number of times }A\text{ occurs}}{\text{number of trials}}}$

Worked example · A biased die

A die is rolled 200 times and a six appears 36 times.

$P(\text{six})\approx\frac{36}{200} =\boxed{0.18}$

This is an experimental estimate. It does not guarantee that the next roll will be a six with probability exactly 0.18.

As the number of trials increases, relative frequency often settles nearer to the underlying probability. Short experiments can vary considerably because of chance.

4. Complementary events

The complement of event $A$ , written $A'$ or “not $A$ ,” consists of all outcomes in which $A$ does not happen.

$\boxed{P(A)+P(A')=1} \qquad\Longrightarrow\qquad \boxed{P(A')=1-P(A)}$

Worked example · On time every day

The probability that Amina is late on at least one day during a week is 0.15. The complementary event is that she is on time every day.

$P(\text{on time every day})=1-0.15 =\boxed{0.85}$

Worked example · At least one friend

The probability that Anu does not meet any friends on the way to school is 0.06. The complement is that she meets at least one friend.

$P(\text{meets at least one friend})=1-0.06 =\boxed{0.94}$

“At least one” is often easiest to calculate by finding the complement: $P(\text{at least one})=1-P(\text{none})$ .

5. Expected number of occurrences

If an event has probability $p$ on each trial and the experiment is repeated $n$ times, the expected number of times the event occurs is:

$\boxed{E(x)=np}$

The expected number is a long-run average, not a promise about the result of one particular set of trials. It can be a decimal even though the actual number of occurrences must be a whole number.

Worked example · Toys in packets

The probability that a packet contains a toy is 0.98. Find the expected number of toys in 30 packets.

$E(x)=30(0.98)=\boxed{29.4\text{ toys}}$

This means that over many groups of 30 packets, the average number of toys per group would be about 29.4. One particular group will contain a whole-number count.

Worked example · Penalty goals

Xenia scores a penalty with probability 0.95. If she takes 10 penalties, her expected number of goals is:

$E(x)=10(0.95)=\boxed{9.5\text{ goals}}$

6. Probability, risk and fair decisions

Probability helps people make decisions when outcomes are uncertain. It is used in areas such as weather forecasting, insurance, medicine and planning. A probability estimate depends on the quality of the data and the assumptions used.

For example, an insurer may use long-term data to estimate the likelihood of an event and plan premiums. A doctor may discuss the probability of benefits and risks of a treatment. In each case, probability informs a decision; it does not tell us exactly what will happen to one individual.

When games of chance involve money, expected values can help reveal how a game is structured. It is important to consider fairness, risk and the possible harm of gambling.

Think critically: Is a probability theoretical or estimated from data? How many trials were used? Does the model reflect the situation fairly?

IB-style practice · Skill questions

A coin is tossed three times. List all possible outcomes. [2 marks]

Show solution

$\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$ .
In a random sample of 40 chocolate boxes, 2 contain a €10 coupon. Use relative frequency to estimate the probability that another box contains a €10 coupon. [2 marks]

Show solution

$P(\text{coupon})\approx\frac{2}{40} =\boxed{0.05}$
In a random sample of 20 students, 8 are basketball players. Estimate the probability that a randomly selected student is a basketball player. [2 marks]

Show solution

$P(\text{basketball player})\approx\frac{8}{20} =\boxed{0.4}$
In a sample of 60 flowers, 6 have six petals. Estimate the probability that a flower selected from the same meadow has six petals. [2 marks]

Show solution

$P(\text{six petals})\approx\frac{6}{60} =\boxed{0.1}$
A biased die is rolled 200 times, and a six appears 36 times. Estimate the probability of rolling a six. [2 marks]

Show solution

$P(\text{six})\approx\frac{36}{200} =\boxed{0.18}$
The probability of rolling an even number on a biased die is $\dfrac{16}{50}$ . Find the probability of rolling an odd number. [2 marks]

Show solution

$P(\text{odd})=1-\frac{16}{50} =\frac{34}{50}=\boxed{0.68}$
Marta has probability 0.87 of cycling to school on at least one day during a particular week. Find the probability that she travels by foot every day that week, assuming these are the only two ways she travels. [2 marks]

Show solution

$P(\text{foot every day})=1-0.87 =\boxed{0.13}$

IB-style practice · Expected outcomes

The probability that a packet of crisps contains a toy is 0.98. Find the expected number of toys in 30 packets. [2 marks]

Show solution

$E(x)=30(0.98)=\boxed{29.4}$
Xenia scores a penalty with probability 0.95. If she takes 10 penalties, find the expected number of goals. [2 marks]

Show solution

$E(x)=10(0.95)=\boxed{9.5\text{ goals}}$
The probability that it snows on any day in Maria Zell is 0.85. Find the expected number of snowy days in December, which has 31 days. [2 marks]

Show solution

$E=31(0.85)=\boxed{26.35\text{ days}}$

This is an expected value, so it need not be a whole number.
A biased coin has probability 0.32 of landing heads. Find the expected number of heads in 30 tosses. [2 marks]

Show solution

$E(x)=30(0.32)=\boxed{9.6\text{ heads}}$
A class has 128 students. The probability that a student is absent on a particular day is 0.1. Find the expected number of absent students. [2 marks]

Show solution

$E(x)=128(0.1)=\boxed{12.8\text{ students}}$

Reflection · The promise of paradise

Jesus’ words to the criminal beside him offer a powerful reminder of grace and hope. Christians look forward to eternal life with Jesus, trusting that his mercy opens the way to be with him. May this hope encourage us to live with faith and kindness while we make wise, thoughtful decisions in the present.

Luke 23:43 (NIV) — “Truly I tell you, today you will be with me in paradise.”

Lesson 4.5 · Probability and expected outcomes

Hope of being with Jesus

Learning objectives

1. Trials, outcomes and events

2. Theoretical probability

3. Experimental probability and relative frequency

4. Complementary events

5. Expected number of occurrences

6. Probability, risk and fair decisions

IB-style practice · Skill questions

IB-style practice · Expected outcomes

Reflection · The promise of paradise

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Probability and expected outcomes AI HL

Lesson 4.5 · Probability and expected outcomes

Hope of being with Jesus

Learning objectives

1. Trials, outcomes and events

2. Theoretical probability

3. Experimental probability and relative frequency

4. Complementary events

5. Expected number of occurrences

6. Probability, risk and fair decisions

IB-style practice · Skill questions

IB-style practice · Expected outcomes

Reflection · The promise of paradise

Related

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Company

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Recommend

alphy school publishers

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