A. Counting and Probability · Permutations and Combinations

Solving Probability Problems with Counting Principles

A2.5 A2.3About 70 minutes

Learning goals

I can solve multi-step probability problems by counting with permutations and combinations.
I can find probabilities for quality-control samples, lotteries, card hands and coin flips.
I can use cases, complements and the additive principle inside probability problems.
I can judge whether outcomes are equally likely before using P(A) = n(A)/n(S).

Words to know

sample (quality control)
A small group of items chosen at random from a larger batch and inspected, so the whole batch need not be tested.

defective
Faulty; not meeting the required standard.

favourable outcome
An outcome in the event you are finding the probability of.

probability distribution (preview)
A list of the possible values of a count (such as the number of heads) with the probability of each. These are the subject of the unit Discrete Probability Distributions.

From puzzles to decisions

A factory cannot test every light bulb it makes, so it tests a small random sample. A lottery corporation must know exactly how likely each prize is before it sets the prize amounts. Card players, sports analysts and medical researchers all ask the same kind of question: out of all the equally likely possibilities, how many give the result I care about?

This lesson puts the whole unit to work: you will choose the right counting tool, split problems into cases or use the complement, and turn the counts into probabilities.

Worked example

Example 1: Quality control

A box of 20 light bulbs contains 3 defective bulbs. An inspector chooses 4 bulbs at random. Find the probability that the sample contains (a) no defective bulbs, (b) exactly one defective bulb, (c) at least one defective bulb.

n(S) = C(20, 4) = 4845 equally likely samples.
(a) All 4 from the 17 good bulbs: C(17, 4) = 2380. P = 2380/4845 ≈ 0.491.
(b) 1 of the 3 defective and 3 of the 17 good: C(3, 1) × C(17, 3) = 3 × 680 = 2040. P = 2040/4845 ≈ 0.421.
(c) Complement of (a): 1 − 2380/4845 = 2465/4845 ≈ 0.509.

(a) ≈ 0.491; (b) ≈ 0.421; (c) ≈ 0.509. The inspector has only about a 51% chance of catching a defect with a sample of 4.

The probabilities for 0, 1, 2 and 3 defective bulbs in the sample. The four bars add to 1 (2380 + 2040 + 408 + 17 = 4845).

Worked example

Example 2: A smaller lottery prize

In Lotto 6/49 you choose 6 different numbers from 1 to 49, and 6 winning numbers are drawn. Find the probability that exactly 3 of your numbers match.

n(S) = C(49, 6) = 13 983 816.
Favourable: 3 of the 6 winning numbers and 3 of the 43 non-winning numbers: C(6, 3) × C(43, 3) = 20 × 12 341 = 246 820.
P = 246 820/13 983 816 ≈ 0.0177, about 1 in 57.

≈ 0.0177 (about 1 chance in 57).

Worked example

Example 3: Coin flips

A fair coin is flipped 8 times. Find (a) P(exactly 3 heads), (b) P(at least 7 heads).

All 2⁸ = 256 sequences are equally likely.
(a) Choose which 3 flips are heads: C(8, 3) = 56. P = 56/256 = 7/32 ≈ 0.219.
(b) At least 7 means 7 or 8 (mutually exclusive cases): C(8, 7) + C(8, 8) = 8 + 1 = 9. P = 9/256 ≈ 0.035.

(a) 7/32 ≈ 0.219; (b) 9/256 ≈ 0.035.

Probabilities for the number of heads in 8 flips. The bar heights follow row 8 of Pascal's triangle (1, 8, 28, 56, 70, 56, 28, 8, 1), divided by 256.

Worked example

Example 4: A random route

A student picks one of the 56 shortest routes to school (5 blocks west, 3 south) completely at random. Find the probability that the route passes the coffee shop at the corner 2 W, 1 S.

From lesson 11, 30 of the 56 routes pass that corner.
Every route is equally likely, so P = 30/56 = 15/28 ≈ 0.536.

15/28 ≈ 0.536

Are the outcomes really equally likely?

P(A) = n(A)/n(S) only works when every outcome in S is equally likely. Every 5-card hand from a shuffled deck is; every route chosen "completely at random" is. But a walker who flips a coin at each corner does not choose all routes equally (forced moves at the edges skip the coin), and the sums of two dice are not equally likely (7 is more likely than 2). When in doubt, build S from outcomes you are sure are equally likely, such as ordered pairs of dice or individual hands.

Worked example

Example 5: Arrangements with identical letters

The letters of BANANA are placed at random in a row. Find the probability that they spell BANANA.

Treat the letters as tiles shuffled at random. There are 6!/(3! × 2!) = 60 distinguishable arrangements, and each is equally likely (each corresponds to the same number, 3! × 2! = 12, of orderings of the labelled tiles).
Exactly one of them is BANANA.

1/60

Just for fun

Flip a coin 10 times. Exactly 5 heads is the single most likely result, yet it happens only about 25% of the time (252/1024). So the most likely thing is still probably not going to happen.

Hands-on activity

Quality control with a deck of cards

You need: 17 number cards (ace to 10, any suits) and 3 face cards, so 20 cards in all. The face cards are the "defective bulbs".

Shuffle the 20 cards well. Deal 4 cards and record the number of face cards (0, 1, 2 or 3). Return the cards and shuffle again.
Repeat for 30 trials. (Working with a partner speeds this up.)
Make a frequency table and draw a bar graph of your results, as a fraction of 30.
Compare your bar graph with the theoretical one in Example 1.
If an inspector wanted at least a 90% chance of finding a defective bulb, would a sample of 4 be enough? Use your results and the theory to explain.

What to look for: expected fractions about 0.49 (0 face cards), 0.42 (1), 0.08 (2) and 0.004 (3), so in 30 trials roughly 15, 13, 2 and 0. Your results will vary. A sample of 4 finds a defect only about 51% of the time, so the inspector needs a much larger sample (with 3 defective in 20, a sample of 11 gives P(at least one) = 1 − C(17, 11)/C(20, 11) ≈ 0.93, while a sample of 10 gives only about 0.89).

Your turn

A box of 10 bulbs has 2 defective bulbs. Three are chosen at random. Find P(no defective bulbs).

In the previous question, find P(at least one defective bulb).

A fair coin is flipped 10 times. Find P(exactly 5 heads).

Find the probability that a 5-card hand contains at least one ace.

The letters of OTTAWA are arranged at random. Find P(the two T's are side by side).

A route 4 blocks east and 2 blocks south is chosen at random from all shortest routes. Find P(it passes the corner 2 E, 1 S).

Find the probability of matching exactly 5 of the 6 winning numbers in Lotto 6/49.

More practice: warm-up

A coin is flipped 3 times. Find P(exactly 2 heads).

A coin is flipped 4 times. Find P(all heads).

Of 4 bulbs, 1 is defective. Two are chosen at random. Find P(the defective bulb is chosen).

Two cards are dealt. Find P(both are aces).

The letters ABCD are arranged at random. Find P(the arrangement starts with A).

More practice: on your own

From a box of 20 bulbs with 3 defective, 4 are chosen. Find P(exactly 2 defective).

A council of 6 is chosen at random from 7 Grade 11 and 8 Grade 12 students. Find P(3 from each grade).

A coin is flipped 8 times. Find P(at least 7 heads).

Find the probability that a 5-card hand is a full house.

The letters of BANANA are arranged at random. Find P(the arrangement spells BANANA).

Challenge

In Lotto 6/49, find the probability of matching exactly 3 of the 6 winning numbers.

A class has 14 boys and 16 girls. Five students are chosen at random. Find P(at least 2 girls).

A committee of 4 is chosen at random from 5 couples (10 people). Find P(no couple is on the committee).

Check your understanding

Quiz

A box of 10 bulbs has 2 defective. Three are chosen at random. What is P(no defective bulbs)?

A fair coin is flipped 8 times. What is P(exactly 3 heads)?

Four bulbs are chosen from 20, of which 3 are defective. Which expression gives P(exactly one defective)?

Two cards are dealt from a standard deck. What is P(both are aces)?

The letters of BANANA are arranged at random. What is P(they spell BANANA)?

Which is usually the quickest way to find P(at least one defective item) in a sample?

For parents and teachers

Big idea (A2.5, A2.3): probability problems become counting problems when the outcomes are equally likely. Choose the counting tool, split into cases or use the complement, and divide by the total. The bar graphs in this lesson preview the hypergeometric and binomial distributions in the unit Discrete Probability Distributions.

How to help: Do the quality-control card activity together; it shows why small samples often miss defects. Ask your teen to explain why the outcomes in each problem are equally likely.

Common difficulties:

Using n(A)/n(S) when outcomes are not equally likely (for example sums of dice).
Forgetting a case in "at least" problems, or not using the complement.
Mixing ordered and unordered counts in the numerator and denominator.
Rounding too early; keep fractions or several digits until the end.