A. Counting and Probability · Permutations and Combinations

Probability Using Counting: Equally Likely Outcomes

A2.5About 70 minutes

Learning goals

I can find a probability by counting favourable and total outcomes with permutations and combinations.
I can solve problems where objects are drawn with replacement and without replacement.
I can check a counting-method probability with a step-by-step (multiplying) method.
I can use the complement to find "at least one" probabilities.

Words to know

equally likely outcomes
Outcomes that each have the same chance of happening, such as the 6 faces of a fair die or every 5-card hand from a well-shuffled deck.

probability of an event
When all outcomes are equally likely, P(A) = n(A) ÷ n(S): the number of favourable outcomes divided by the total number of outcomes.

with replacement
Each object is put back before the next draw, so the draws do not affect each other.

without replacement
Objects are not put back, so each draw changes what is left for the next draw.

complement rule
P(not A) = 1 − P(A).

Counting meets probability

**Remember from the unit Probability: when all outcomes in a sample space S are equally likely**, the probability of an event A is

P(A) = n(A) / n(S)

That formula is easy when you can list the outcomes. But how likely is a 5-card hand to be all hearts? There are 2 598 960 hands. Nobody is going to list them. With combinations, you do not have to: n(S) = C(52, 5) and n(A) = C(13, 5).

This lesson uses everything from the unit so far to find probabilities by counting.

The marble problem

The curriculum asks: two marbles are drawn at random from a bag of 12 green and 16 red marbles; what is the probability that both are green (a) if the first marble is replaced, (b) if it is not replaced? We will use 12 green and 16 blue marbles so the picture's colours match; the mathematics is identical.

A bag of 28 marbles: 12 green and 16 blue.

Worked example

Example 1: With replacement

Draw a marble, note its colour, put it back, mix, and draw again. Find P(both green).

Count ordered outcomes. Each draw can be any of the 28 marbles: n(S) = 28 × 28 = 784.
Favourable: green then green, 12 × 12 = 144.
P = 144/784 = 9/49 ≈ 0.184.
Check by multiplying probabilities: (12/28) × (12/28) = (3/7)² = 9/49 ✓.

9/49 ≈ 0.184

Worked example

Example 2: Without replacement

Draw two marbles at once (or one after the other without putting the first back). Find P(both green).

The order of the two marbles does not matter, so count combinations: n(S) = C(28, 2) = 378.
Favourable: choose 2 of the 12 green marbles: C(12, 2) = 66.
P = 66/378 = 11/63 ≈ 0.175.
Check step by step: first green 12/28, then 11 green out of 27 left: (12/28) × (11/27) = 132/756 = 11/63 ✓.

11/63 ≈ 0.175. It is a bit smaller than 9/49 because once a green marble is removed, fewer greens are left.

The only difference is the second branch: 12/28 when the first marble goes back, 11/27 when it does not.

Count the same way on top and bottom

If you count ordered outcomes for n(S) (permutations, or multiplying draws), count ordered outcomes for n(A) too. If you count unordered selections (combinations) for n(S), use combinations for n(A). Mixing the two is the most common error in this lesson. Either way gives the same probability.

Worked example

Example 3: Arrangements

The letters of CANOES are arranged at random. What is the probability that the three vowels are together?

All 6! = 720 arrangements are equally likely.
From lesson 5, the vowels are together in 4! × 3! = 144 arrangements.
P = 144/720 = 1/5.

1/5 = 0.2

Worked example

Example 4: Committees and complements

A 5-person committee is chosen at random from 6 teachers and 8 students. Find (a) P(exactly 2 teachers), (b) P(at least one teacher).

n(S) = C(14, 5) = 2002.
(a) n(A) = C(6, 2) × C(8, 3) = 15 × 56 = 840. P = 840/2002 = 60/143 ≈ 0.420.
(b) Complement: P(no teachers) = C(8, 5)/2002 = 56/2002 = 4/143. So P(at least one) = 1 − 4/143 = 139/143 ≈ 0.972.

(a) 60/143 ≈ 0.420; (b) 139/143 ≈ 0.972.

Worked example

Example 5: A hand of cards

Five cards are dealt from a well-shuffled standard deck. What is the probability that all five are hearts?

n(S) = C(52, 5) = 2 598 960.
n(A) = C(13, 5) = 1287.
P = 1287/2 598 960 ≈ 0.000 495, or about 1 in 2020.

about 0.000 495

Just for fun

Only 4 of the 2 598 960 possible five-card hands are royal flushes (10, jack, queen, king and ace of one suit). That is about 1 chance in 650 000 per deal, so do not plan your evening around it.

Hands-on activity

Card draws: with and without replacement

You need: 3 red cards and 2 black cards from a deck (5 cards), and a paper cup or hat to draw from.

Calculate P(both red) for drawing 2 cards with replacement and without replacement.
Without replacement: shuffle, draw 2 cards, record whether both are red, return them. Repeat 40 times.
With replacement: draw 1 card, record, put it back, shuffle, draw again. Repeat 40 times.
Compare your experimental fractions with the theoretical values. Which method gave more "both red" results?

What to look for: without replacement, C(3, 2)/C(5, 2) = 3/10 = 0.30; with replacement, (3/5)² = 9/25 = 0.36. In 40 trials expect roughly 12 and 14 "both red" results; small samples vary a lot, so do not be surprised by results like 9 or 17. Pooling results with someone else gives a better estimate (the unit Probability showed that experimental probability approaches theoretical probability as trials increase).

Your turn

A bag holds 5 green and 3 blue marbles. Two are drawn without replacement. Find P(both green).

Same bag, but the first marble is replaced before the second draw. Find P(both green).

The letters of WINTER are arranged at random. Find P(the arrangement begins with W).

Find the probability that a 5-card hand contains only hearts.

A 4-digit PIN is chosen at random (repeats allowed). Find P(no digit is repeated).

Three boys and two girls line up at random. Find P(a girl is at each end).

What is the probability of matching all 6 winning numbers in the Lotto 6/49 Classic Draw with one selection?

More practice: warm-up

Two dice are rolled. Find P(sum of 7).

One of 10 people is chosen at random; 4 are in Grade 12. Find P(a Grade 12 is chosen).

From 3 green and 2 blue marbles, two are drawn with replacement. Find P(both green).

Same bag, without replacement. Find P(both green).

The letters A, B, C are arranged at random. Find P(they are in alphabetical order).

More practice: on your own

A 3-person committee is chosen at random from 4 boys and 5 girls. Find P(all girls).

For the same committee, find P(at least one boy).

Six people sit in a row at random. Find P(two particular friends sit together).

Find P(a 5-card hand contains exactly 2 aces).

The letters of TORONTO are arranged at random. Find P(the arrangement begins with T).

Challenge

Find the probability that at least two of 23 people share a birthday. Assume 365 equally likely birthdays and ignore February 29.

Three marbles are drawn without replacement from 12 green and 16 blue. Find P(at least one green).

Five couples sit at random in a row of 10 seats. Find P(one particular couple sits together).

Check your understanding

Quiz

A bag has 12 green and 16 blue marbles. Two are drawn with replacement. What is P(both green)?

Same bag, without replacement. What is P(both green)?

Why is P(both green) smaller without replacement?

Six people sit in a row at random. What is the probability that two particular friends sit together?

When may you use P(A) = n(A)/n(S)?

Which expression gives the probability that a 5-card hand is all hearts?

For parents and teachers

Big idea (A2.5): for equally likely outcomes, a probability is a ratio of two counts, and permutations and combinations find those counts. With replacement, draws are independent; without replacement, each draw changes the next. Counting and multiplying probabilities along branches give the same answer.

How to help: Do the card activity together and compare the results with the theory. Ask your teen to solve one problem both ways (counting, and multiplying along branches).

Common difficulties:

Counting ordered outcomes on top and unordered on the bottom (or the reverse).
Treating "without replacement" as if the draws were independent.
Forgetting that "at least one" is usually easiest as 1 − P(none).
Reporting probabilities greater than 1, which signals a counting error.