A. Counting and Probability · Probability

Solving Multi-Step Probability Problems

A1.5 A1.6About 75 minutes

Learning goals

I can choose an efficient strategy (Venn diagram, tree diagram, two-way table, list or formula) for a probability problem.
I can find the total probability of an event that can happen along several branches of a tree.
I can "reverse" a conditional probability, finding P(A | B) from P(B | A), using a tree or a table of natural frequencies.
I can explain why a positive result from an accurate test can still be wrong much of the time when the condition is rare.

Words to know

total probability
If an event B can happen along several mutually exclusive paths, P(B) is the sum of the path probabilities: P(B) = P(A)P(B | A) + P(A′)P(B | A′).

natural frequencies
Counts out of an imaginary group (such as 1000 people) used in place of probabilities to make multi-step problems easier to follow.

false positive
A test result that says "yes" when the true answer is "no".

false negative
A test result that says "no" when the true answer is "yes".

Bayes' theorem
A formula for reversing a conditional probability: P(A | B) = P(A)P(B | A) / P(B). You can always get the same answer from a tree diagram.

Putting the tools together

You now have a full toolkit: sample spaces, complements, Venn diagrams, the addition rules, the multiplication rules and conditional probability. Real problems usually need several of them. Choose the tool that fits the problem:

Overlapping categories ("plays hockey or soccer"): Venn diagram.
Steps in sequence (draws, games, stages of production): tree diagram.
Data counted two ways: two-way table.
Many independent trials and "at least one": complement and multiplication.
"Given" information that is backwards from what you want: tree diagram, then the conditional probability formula.

In every case, write down what each event means, in words, before calculating.

Worked example

Example 1: Total probability

A factory has two machines. Machine A makes 60% of the light bulbs and 2% of its bulbs are defective. Machine B makes 40% and 5% of its bulbs are defective. A bulb is chosen at random. (a) Find P(defective). (b) Given that the bulb is defective, find P(it came from machine A).

Draw a tree: first the machine (A: 0.6, B: 0.4), then defective D or OK.
P(A ∩ D) = 0.6 × 0.02 = 0.012. P(B ∩ D) = 0.4 × 0.05 = 0.020.
(a) A defective bulb comes from A or B (mutually exclusive paths), so add: P(D) = 0.012 + 0.020 = 0.032.
(b) P(A | D) = P(A ∩ D) / P(D) = 0.012 / 0.032 = 0.375.

(a) P(defective) = 0.032. (b) P(A | defective) = 0.375. Machine A makes most of the bulbs, but only 37.5% of the defective ones.

Tree diagram for Example 1. The two paths ending in D (defective) add to 0.032. Of that, machine A's path contributes 0.012, so P(A | D) = 0.012/0.032 = 0.375.

Worked example

Example 2: Reversing a conditional with natural frequencies

About 30% of the email sent to a school account is spam. A filter flags 96% of spam messages, but it also flags 2% of real messages by mistake. (a) What fraction of emails is flagged? (b) If an email is flagged, what is the probability it is really spam? (c) If an email is not flagged, what is the probability it is spam?

Imagine 1000 emails. 300 are spam and 700 are not.
Spam that is flagged: 96% of 300 = 288. Spam not flagged: 12.
Real email flagged by mistake: 2% of 700 = 14. Real email not flagged: 686.
(a) Flagged: 288 + 14 = 302 out of 1000, so P(flagged) = 0.302.
(b) Of the 302 flagged emails, 288 are spam: P(spam | flagged) = 288/302 ≈ 0.954.
(c) Not flagged: 12 + 686 = 698. Of these, 12 are spam: P(spam | not flagged) = 12/698 ≈ 0.017.

(a) 0.302, (b) ≈ 0.954, (c) ≈ 0.017. Natural frequencies give the same answers as the formulas, with numbers that are easier to picture.

Natural frequency tree for 1000 emails. To find P(spam | flagged), compare the flagged spam (288) with all flagged emails (288 + 14).

Worked example

Example 3: A two-day weather chain

In a northern Ontario town in winter, P(snow today) = 0.4. If it snows today, P(snow tomorrow) = 0.6. If it does not snow today, P(snow tomorrow) = 0.2. (Example values.) Find (a) P(snow tomorrow), (b) P(snow on at least one of the two days), (c) P(snow today | snow tomorrow).

Let T = snow today and M = snow tomorrow. P(T) = 0.4, P(T′) = 0.6, P(M | T) = 0.6, P(M | T′) = 0.2.
(a) Total probability: P(M) = 0.4 × 0.6 + 0.6 × 0.2 = 0.24 + 0.12 = 0.36.
(b) Complement: P(no snow either day) = P(T′) × P(M′ | T′) = 0.6 × 0.8 = 0.48. So P(at least one) = 1 − 0.48 = 0.52.
(c) P(T | M) = P(T ∩ M)/P(M) = 0.24/0.36 = 2/3.

(a) 0.36, (b) 0.52, (c) 2/3. Snowy days tend to come in runs, so these events are dependent: P(M | T) = 0.6 is not equal to P(M) = 0.36.

A checklist for multi-step problems

Name the events with letters and words.
Write what you are given in probability notation. Watch for conditionals: "of those who…", "if…", "given…".
Pick a tool: tree, table, Venn or formula. If the given conditionals are the reverse of what is asked, use a tree or natural frequencies.
Calculate, keeping fractions or at least 4 decimal places until the end.
Check: probabilities between 0 and 1, all tree paths add to 1, and the answer makes sense in context.

Why rare conditions fool accurate tests

Suppose a screening test is very accurate, but the condition it looks for is rare. Most people tested do not have the condition, so even a small false-positive rate produces many false positives. They can outnumber the true positives. That is why a positive screening result is usually followed by a second, more specific test.

The Challenge questions below explore this. The same idea matters for spam filters, fraud detection, security screening, and any situation where you search for something rare.

Just for fun

Probability puzzle humour: a statistician gets a positive result on a test that is "99% accurate" and calmly asks, "Great, but what's the base rate?" Everyone else in the waiting room is very confused.

Hands-on activity

Natural frequency challenge

You need: paper, a pencil, a calculator, and 100 small items such as dried beans, buttons or pasta pieces (or just draw 100 dots in a 10 × 10 grid).

Each item stands for one home in a town of 100 homes. In this town, 10% of homes have a certain leaky faucet. A plumber's quick check spots 80% of leaky faucets and wrongly reports a leak in 10% of homes without one.
Separate 10 items as "leaky" homes and 90 as "not leaky".
Of the 10 leaky homes, move 8 to a "reported leak" pile. Of the 90 others, move 9 to the "reported leak" pile.
Count the "reported leak" pile. What fraction of it really leaks?
Compare with the formula: P(leaky | reported) = (0.1 × 0.8) / (0.1 × 0.8 + 0.9 × 0.1).

What to look for: the reported pile has 8 + 9 = 17 homes, and only 8 really leak, so P(leaky | reported) = 8/17 ≈ 0.47. The formula gives 0.08/0.17 ≈ 0.47 too. Even with a fairly good check, more than half of the "leak" reports are wrong, because leaky homes are rare. Keep small items away from young children.

Your turn

Two machines make phone cases: X makes 70% (3% defective) and Y makes 30% (6% defective). Find P(a case is defective).

Using the phone case question, find P(made by Y | defective).

Bag 1 has 3 red and 1 blue marble. Bag 2 has 1 red and 3 blue. A coin toss picks the bag, then one marble is drawn. Find P(red).

Using the two bags, find P(bag 1 | red).

A student is on time with probability 0.9 if the bus is on time and 0.5 if the bus is late. The bus is late 20% of the time. Find P(student is on time).

Using the bus question, find P(bus was late | student was late).

More practice: warm-up

On a tree diagram, what operation gives the probability of one path?

How do you combine paths that all lead to the same event?

P(A) = 0.3, P(B | A) = 0.5. P(A ∩ B)?

Out of 1000 people, 20 have a condition. Write this as a probability.

What is a false positive?

More practice: on your own

A basketball player makes her first free throw with probability 0.8. If she makes the first, she makes the second with probability 0.85; if she misses the first, she makes the second with probability 0.6. Find P(she makes exactly one).

Using the free-throw question, find P(she made the first | she made exactly one).

In a school, 55% of students are in the Grade 12 university stream. Of those, 70% take MDM4U; of the others, 20% take MDM4U. Find P(a student takes MDM4U) and P(university stream | takes MDM4U).

Two cards are drawn without replacement. Find P(the second card is an ace).

A team wins 70% of home games and 45% of away games. Half its games are at home. Given that it won a game, find P(the game was at home).

Challenge

A condition affects 1% of a population. A screening test detects it 95% of the time when it is present, and gives a false positive 8% of the time when it is absent. Find P(condition | positive test). Use 10 000 people.

For the test in the previous question, find P(no condition | negative test). Why is this reassuring?

A best-of-five series ends when a team wins 3 games. Team A wins each game with probability 0.6, independently. Find P(A wins the series in exactly 4 games).

Check your understanding

Quiz

Machine A makes 50% of a product with a 4% defect rate; machine B makes 50% with a 2% defect rate. What is P(defective)?

Using the previous question, what is P(machine A | defective)?

Which tool is usually best when the given conditional probabilities are the reverse of what you are asked to find?

P(snow today) = 0.5. P(snow tomorrow | snow today) = 0.7 and P(snow tomorrow | no snow today) = 0.3. What is P(snow tomorrow)?

A rare condition (1 in 1000) is screened with a test that has a 5% false-positive rate. Why are most positive results false?

In a tree diagram, all of the end-of-path probabilities should add to:

For parents and teachers

Big idea (A1.5 and A1.6 together): harder problems combine several rules. Trees organize sequences; adding the matching paths gives total probabilities; dividing a path by the total for a condition reverses a conditional probability. Thinking in "natural frequencies" (out of 1000 people) makes these problems much easier to follow.

How to help: Ask your teen to explain Example 2 using the numbers out of 1000 emails, without any formulas. If they can explain why most positive results of a rare-condition test can be false, they understand this lesson. The bean or button activity makes it concrete.

Common difficulties:

Confusing P(positive | condition) with P(condition | positive).
Forgetting one of the paths when finding a total probability.
Rounding too early in multi-step calculations.
Not naming events clearly before calculating.

The medical-test challenge is about how screening works mathematically; it is not health advice.