NSWMaths Extension 1Syllabus dot point

When every outcome is equally likely, how do you turn a probability into a ratio of two counts, and how do the counting moves (multiplication principle, permutations, combinations, complement, cases) supply both the favourable count and the total?

Compute probabilities of equally likely outcomes as P = (favourable outcomes)/(total outcomes) where both counts come from the counting methods (multiplication principle, ^{n}P_{r} and ^{n}C_{r}), and apply the complement, separate-pool multiplication, cases and inclusion-exclusion to find the favourable count for selection, card-hand, digit-string and with/without-replacement problems

Turning counting into probability. When outcomes are equally likely, P = favourable/total and both counts come from the multiplication principle, nPr and nCr. Committee and card-hand probabilities by combinations, exactly-k of a type, at-least-one by the complement, a five-digit-number probability, inclusion-exclusion, and with versus without replacement.

Generated by Claude Opus 4.819 min answerUpdated 2026-06-21

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

For equally likely outcomes, $P(E) = \dfrac{n(E)}{n(S)} = \dfrac{\text{favourable}}{\text{total}}$ , and both counts come from the counting methods. Decide ordered or unordered once and count both parts that way (the common factor cancels); for selections use combinations. A committee of $4$ from $7$ women and $5$ men has $^{12}C_{4} = 495$ outcomes, so $P(\text{all women}) = \dfrac{^{7}C_{4}}{^{12}C_{4}} = \dfrac{35}{495} = \dfrac{7}{99}$ and $P(\text{exactly } 2 \text{ men}) = \dfrac{^{5}C_{2}\,^{7}C_{2}}{^{12}C_{4}} = \dfrac{14}{33}$ (multiply separate pools). A $5$ -card hand has $^{52}C_{5} = 2\,598\,960$ outcomes; $P(\text{exactly } 2 \text{ hearts}) = \dfrac{^{13}C_{2}\,^{39}C_{3}}{^{52}C_{5}} = \dfrac{9139}{33\,320}$ . Handle "at least one" by the complement: $P(\text{at least one ace}) = 1 - \dfrac{^{48}C_{5}}{^{52}C_{5}} = \dfrac{18\,472}{54\,145}$ . For a five-digit number ( $9 \times 10^4 = 90\,000$ of them), $P(\text{at least one } 4) = \dfrac{90\,000 - 8 \times 9^4}{90\,000} = \dfrac{521}{1250}$ , and "at least one $4$ and one $5$ " needs inclusion-exclusion, giving $\dfrac{856}{5625}$ . With replacement the draws are independent and multiply, $\left(\dfrac{7}{15}\right)^3$ , while without replacement it is a ratio of combinations, $\dfrac{^{7}C_{3}}{^{15}C_{3}} = \dfrac{1}{13}$ .

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What this dot point is asking
The answer
How exam questions ask about this dot point

What this dot point is asking

The three counting pages of this module gave you machinery for counting outcomes: the multiplication principle and ordered selections for counting in stages with $n^r$ and $^{n}P_{r}$ , the grouping, complement and cases moves for restricted arrangements, and combinations for unordered selections with $^{n}C_{r}$ . This page puts that machinery to work on probability. NESA's ME-A1 dot point asks you to compute a probability by counting: find the number of favourable outcomes and the number of total outcomes, then divide.

The single idea is the equally likely model. When every outcome of an experiment is equally likely, the probability of an event $E$ is

P(E) = \frac{\text{number of outcomes in } E}{\text{number of outcomes in the sample space}} = \frac{n(E)}{n(S)},

and the whole skill is using counting to fill in those two numbers. Both the numerator and the denominator are counts of the same kind of object, so they must be counted the same way: if the sample space is "all $5$ -card hands", the favourable count must also be a count of hands. Everything else is choosing the right counting tool for each count, the multiplication principle, a permutation, a combination, the complement, cases, or inclusion-exclusion, and being consistent about ordered versus unordered between the two counts.

The answer

The equally likely model: probability as a ratio of two counts

A probability experiment has a set of possible outcomes called the sample space $S$ , and an event $E$ is a subset of those outcomes. When the outcomes are equally likely, no single outcome is favoured over another, so the probability of $E$ is simply the share of outcomes that lie in $E$ . Drawing a card "at random", dealing a hand "at random", choosing a committee "at random", arranging letters "in random order": each phrase signals that the underlying outcomes are equally likely, which is exactly the licence to count.

The reason the formula works is a direct sum. Each of the $n(S)$ outcomes carries the same probability $\dfrac{1}{n(S)}$ , and the event $E$ contains $n(E)$ of them, so

P(E) = \underbrace{\frac{1}{n(S)} + \frac{1}{n(S)} + \cdots + \frac{1}{n(S)}}_{n(E) \text{ terms}} = \frac{n(E)}{n(S)}.

So a probability question becomes two counting questions: count the sample space, and count the favourable event. The art is choosing what an "outcome" is and counting consistently.

Ordered or unordered? Choose once, and use it for both counts

Before counting anything, settle the two questions that run through all of combinatorics: are the outcomes ordered or unordered, and, if ordered, is repetition allowed? The answer decides which tool counts the sample space, and you must then count the favourable event the same way. Both choices are valid as long as they match, because any consistent factor cancels in the ratio.

A clean illustration is dealing $3$ cards from a pack and asking for the probability of one club then two hearts in that order. Counting ordered outcomes, the sample space is $^{52}P_{3} = 52 \times 51 \times 50 = 132\,600$ , and the favourable outcomes are club, heart, heart, which is $13 \times 13 \times 12 = 2028$ , giving $\dfrac{2028}{132\,600} = \dfrac{13}{850}$ . If instead the question asked for one club and two hearts in any order, the natural count is unordered: the sample space is $^{52}C_{3} = 22\,100$ and the favourable count is $^{13}C_{1}\,^{13}C_{2} = 13 \times 78 = 1014$ , giving $\dfrac{1014}{22\,100} = \dfrac{39}{850}$ , three times larger because the club can sit in any of three positions. The rule of thumb: if a problem can be done either way, use unordered selections, because the numbers are smaller and the combinations are easier to handle.

Exam technique

Markers reward a clearly identified sample space and a matching favourable count. Lock in the marks like this:

Write the sample space size first, and say what an outcome is. "All $5$ -card hands, $^{52}C_{5}$ " or "all five-digit numbers, $9 \times 10^4$ ". That one line frames everything and usually secures a method mark.
Decide ordered versus unordered once, and count both parts that way. A ratio of two unordered counts and a ratio of two ordered counts give the same probability; mixing them is the classic error.
Default to combinations for selections, hands and committees ("chosen at random", "dealt", "drawn together"); the numbers are smaller than the ordered version.
Leave the fraction unsimplified until the end, then reduce and (if asked) give a decimal. Keep one source of truth: compute a count such as $^{52}C_{5} = 2\,598\,960$ once and reuse it in every part.
Read "at least", "at most" and "not" as complement cues, and read "or" as a possible inclusion-exclusion: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .

Selection probabilities with combinations

The commonest equally likely model is a random selection: a committee, a team, a hand, a handful of discs. Order does not matter, so the sample space is a combination, and the favourable count is a combination (or a product of combinations from separate pools). The whole pattern is

P(\text{event}) = \frac{\text{combinations that make the event}}{\text{all combinations of that size}}.

Take a committee of $4$ chosen at random from $7$ women and $5$ men, $12$ people in all. The sample space is every committee of $4$ , namely $^{12}C_{4} = 495$ equally likely committees. To find $P(\text{all } 4 \text{ are women})$ , count the all-women committees, $^{7}C_{4} = 35$ , and divide:

P(\text{all women}) = \frac{^{7}C_{4}}{^{12}C_{4}} = \frac{35}{495} = \frac{7}{99} \approx 0.0707.

The same total $495$ serves every event for this committee. The diagram below shows how the $495$ committees split by the number of men, and where the favourable count of $35$ sits.

The bars make the shape of the problem visible: most committees are mixed, the all-women and all-men committees are rare, and the five case counts $35 + 175 + 210 + 70 + 5 = 495$ recover the whole sample space. That last fact is a free check: the favourable counts of all the disjoint cases must add to the total.

Exactly $k$ of a type, and the separate-pools rule

When an event fixes how many come from each pool, choose from each pool separately and multiply the combinations, then divide by the total. Staying with the committee of $4$ from $7$ women and $5$ men, the probability of exactly $2$ men chooses $2$ of the $5$ men and $2$ of the $7$ women:

P(\text{exactly } 2 \text{ men}) = \frac{^{5}C_{2}\,^{7}C_{2}}{^{12}C_{4}} = \frac{10 \times 21}{495} = \frac{210}{495} = \frac{14}{33} \approx 0.4242.

The numerator multiplies because the two pools are chosen independently; the denominator is the same $495$ . Running through every value of "number of men" gives the full distribution, and the favourable counts $35, 175, 210, 70, 5$ are exactly the column counts in the diagram above. Because these cases are disjoint and exhaustive, their probabilities sum to $1$ :

\frac{35 + 175 + 210 + 70 + 5}{495} = \frac{495}{495} = 1,

which is the most reliable check available on a "by cases" probability.

Card-hand probabilities

A standard pack has $52$ cards in $4$ suits of $13$ . A poker-style hand of $5$ cards is an unordered selection, so the sample space is

^{52}C_{5} = 2\,598\,960

equally likely hands, a number worth computing once and reusing. Every hand probability is then a combination count over $2\,598\,960$ . Three patterns cover most questions.

Exactly $r$ of one suit. For exactly $2$ hearts, choose $2$ of the $13$ hearts and the other $3$ from the $39$ non-hearts:

P(\text{exactly } 2 \text{ hearts}) = \frac{^{13}C_{2}\,^{39}C_{3}}{^{52}C_{5}} = \frac{78 \times 9139}{2\,598\,960} = \frac{712\,842}{2\,598\,960} = \frac{9139}{33\,320} \approx 0.2743.

A specified split across two suits. For $3$ spades and $2$ hearts, choose from each suit and multiply:

P(3\spadesuit, 2\heartsuit) = \frac{^{13}C_{3}\,^{13}C_{2}}{^{52}C_{5}} = \frac{286 \times 78}{2\,598\,960} = \frac{22\,308}{2\,598\,960} = \frac{143}{16\,660} \approx 0.0086.

All five of one suit (a flush). Any one of the $4$ suits supplies all $5$ cards, so multiply by $4$ :

P(\text{all one suit}) = \frac{4 \times {}^{13}C_{5}}{^{52}C_{5}} = \frac{4 \times 1287}{2\,598\,960} = \frac{5148}{2\,598\,960} = \frac{33}{16\,660} \approx 0.0020.

Each is favourable-over-total with both counts as combinations of $5$ cards, the only difference being how the favourable hands are built.

"At least one" by the complement

"At least one" almost always fragments into several cases ("exactly one", "exactly two", ...), so it is far shorter to count the complement, "none", and subtract its probability from $1$ :

P(\text{at least one}) = 1 - P(\text{none}).

Deal $5$ cards and ask for $P(\text{at least one ace})$ . The complement is a hand with no ace, drawn from the $48$ non-aces, $^{48}C_{5} = 1\,712\,304$ , so

P(\text{no ace}) = \frac{^{48}C_{5}}{^{52}C_{5}} = \frac{1\,712\,304}{2\,598\,960} = \frac{35\,673}{54\,145} \approx 0.6588,

P(\text{at least one ace}) = 1 - \frac{35\,673}{54\,145} = \frac{18\,472}{54\,145} \approx 0.3412.

Counting "at least one ace" directly would mean adding the exactly-one, exactly-two, exactly-three and exactly-four ace cases; the complement collapses all of that into one clean count.

A five-digit-number probability, and inclusion-exclusion

Digit problems use the multiplication principle for both counts. A five-digit number chosen at random has a first digit from $1$ to $9$ ( $9$ choices, since a number cannot start with $0$ ) and four further digits each from $0$ to $9$ , so the sample space is

9 \times 10^4 = 90\,000

equally likely numbers. For $P(\text{at least one } 4)$ , reach for the complement: a number with no $4$ has $8$ choices for the first digit ( $1$ to $9$ except $4$ ) and $9$ for each of the other four ( $0$ to $9$ except $4$ ), so $8 \times 9^4 = 52\,488$ have no $4$ , and the rest have at least one:

P(\text{at least one } 4) = \frac{90\,000 - 52\,488}{90\,000} = \frac{37\,512}{90\,000} = \frac{521}{1250} \approx 0.4168.

The case tree shows the split: the whole sample space branches into "no $4$ " and "at least one $4$ ", and the favourable count is what is left after removing the complement.

When two "at least one" conditions are joined by "and", the complement of the event is an "or", and an "or" of overlapping sets needs inclusion-exclusion. For $P(\text{at least one } 4 \textbf{ and } \text{at least one } 5)$ , let $A$ be the numbers with no $4$ and $B$ the numbers with no $5$ . Then the numbers failing the event (no $4$ or no $5$ ) form $A \cup B$ , and

n(A \cup B) = n(A) + n(B) - n(A \cap B).

Here $n(A) = n(B) = 8 \times 9^4 = 52\,488$ , and $A \cap B$ (no $4$ and no $5$ ) has $7$ choices for the first digit and $8$ for each of the rest, $7 \times 8^4 = 28\,672$ . So

n(A \cup B) = 52\,488 + 52\,488 - 28\,672 = 76\,304,

and the favourable numbers (at least one $4$ and at least one $5$ ) are everything else:

P(\ge 1 \text{ four and} \ge 1 \text{ five}) = \frac{90\,000 - 76\,304}{90\,000} = \frac{13\,696}{90\,000} = \frac{856}{5625} \approx 0.1522.

The key move is recognising that "and" on two at-least-one conditions complements to an "or", and that the "or" overlaps, so the overlap $A \cap B$ must be subtracted once.

With replacement versus without replacement

Whether an item is replaced between draws changes the model, and so changes both counts. Without replacement (or drawing all at once), the items drawn are distinct, so a selection of $r$ is naturally an unordered combination, and the probability is a ratio of combinations. With replacement, each draw is from the full set and the draws are independent, so probabilities multiply.

Take a bag of $3$ red, $7$ yellow and $5$ blue balls ( $15$ in all) and draw $3$ . Without replacement, the sample space is $^{15}C_{3} = 455$ and $P(\text{all yellow})$ uses the favourable count $^{7}C_{3} = 35$ :

P(\text{all yellow, no replacement}) = \frac{^{7}C_{3}}{^{15}C_{3}} = \frac{35}{455} = \frac{1}{13} \approx 0.0769.

With replacement, each draw is yellow with probability $\dfrac{7}{15}$ independently, so

P(\text{all yellow, with replacement}) = \left(\frac{7}{15}\right)^3 = \frac{343}{3375} \approx 0.1016.

Replacement is slightly more likely here, because the proportion of yellow is held at $\dfrac{7}{15}$ on every draw, whereas without replacement each yellow removed leaves a smaller proportion for the next draw. Spotting "drawn together" or "all at once" (no replacement, combinations) versus "replaced before the next" (with replacement, independent multiplication) is half the battle on these questions.

Key fact

Choosing the model.

Equally likely selection (without replacement / all at once): sample space and event are both combinations, $P = \dfrac{\text{favourable combinations}}{\text{total combinations}}$ .
Ordered with repetition (digit strings, with replacement): count both parts by the multiplication principle / powers, or multiply independent per-draw probabilities.
At least one: $P(\text{at least one}) = 1 - P(\text{none})$ .
An "or" of overlapping events: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .

Count the numerator and denominator the same way, and the consistent factor cancels.

How exam questions ask about this dot point

The wording tells you the model and the move. Map the phrase to the method:

"...chosen / selected / drawn at random", "a committee / team / hand is formed". Equally likely outcomes: $P = \dfrac{\text{favourable}}{\text{total}}$ , almost always with combinations.
"...all of one type", "all male", "all the same suit". Favourable is a single combination from one pool over the total: $\dfrac{^{a}C_{r}}{^{n}C_{r}}$ .
"...exactly $k$ of one type and the rest another", "two men and three women". Multiply per-pool combinations in the numerator: $\dfrac{^{a}C_{i}\,^{b}C_{j}}{^{n}C_{r}}$ .
"...at least one", "at most", "none ... so find the chance of some". Complement: $1 - P(\text{none})$ .
"...at least one $X$ and at least one $Y$ ", "contains both". Complement to an "or", then inclusion-exclusion on the overlap.
"...a majority", "more Labor than Liberal", "at least $3$ ". Split into the qualifying cases, count each by multiplying pools, then add.
"...a five-digit number / a four-digit code is chosen at random; find the probability that ...". Multiplication principle for both counts; mind the no-leading-zero restriction.
"...drawn together / all at once / without replacement" versus "...replaced before the next draw". Combinations for without replacement; independent multiplication for with replacement.

Worked examples

Find a committee probability for all-of-one-type and exactly-k

A committee of $4$ is chosen at random from $7$ women and $5$ men. Find the probability that (a) all four are women, (b) exactly two are men, and (c) check that the probabilities over all possible numbers of men add to $1$ .

Set up the sample space. A committee is an unordered selection of $4$ from the $12$ people, so

n(S) = {}^{12}C_{4} = 495 \text{ equally likely committees.}

Part (a): count the all-women committees. All $4$ come from the $7$ women, $^{7}C_{4} = 35$ , so

P(\text{all women}) = \frac{^{7}C_{4}}{^{12}C_{4}} = \frac{35}{495} = \frac{7}{99} \approx 0.0707.

Part (b): exactly two men, multiplying the two pools. Choose $2$ of the $5$ men and $2$ of the $7$ women:

P(\text{exactly } 2 \text{ men}) = \frac{^{5}C_{2}\,^{7}C_{2}}{^{12}C_{4}} = \frac{10 \times 21}{495} = \frac{210}{495} = \frac{14}{33} \approx 0.4242.

Part (c): the cases must exhaust the sample space. The favourable counts for $0, 1, 2, 3, 4$ men are

^{7}C_{4}, \ ^{5}C_{1}\,^{7}C_{3}, \ ^{5}C_{2}\,^{7}C_{2}, \ ^{5}C_{3}\,^{7}C_{1}, \ ^{5}C_{4} = 35, 175, 210, 70, 5,

which add to $35 + 175 + 210 + 70 + 5 = 495$ , so the probabilities sum to $\dfrac{495}{495} = 1$ , as they must.

Final answer. (a) $\dfrac{7}{99} \approx 0.0707$ ; (b) $\dfrac{14}{33} \approx 0.4242$ ; (c) the five case probabilities total $1$ .

Compute a card-hand probability two consistent ways

Three cards are dealt from a standard pack of $52$ . Find the probability that the hand is one club and two hearts, in any order. Confirm the answer by counting ordered outcomes instead.

Unordered count: choose the suits separately. The sample space is every $3$ -card hand, $^{52}C_{3} = 22\,100$ . The favourable hands choose $1$ of the $13$ clubs and $2$ of the $13$ hearts:

^{13}C_{1}\,^{13}C_{2} = 13 \times 78 = 1014,

P(1 \text{ club}, 2 \text{ hearts}) = \frac{1014}{22\,100} = \frac{39}{850} \approx 0.0459.

Ordered check: count arrangements instead. Counting ordered deals, the sample space is $^{52}P_{3} = 132\,600$ . The favourable ordered deals must still be one club and two hearts in some order. The club can take any of the $3$ positions ( $3$ ways), the club itself is one of $13$ , and the two hearts fill the other two positions in $13 \times 12$ ordered ways:

3 \times 13 \times (13 \times 12) = 3 \times 13 \times 156 = 6084.

Then

P = \frac{6084}{132\,600} = \frac{39}{850} \approx 0.0459,

the same value. The ordered numerator and denominator are each $6$ times the unordered ones (the $3! = 6$ orders of three cards), and the factor cancels.

Final answer. $P(1 \text{ club}, 2 \text{ hearts}) = \dfrac{39}{850} \approx 0.0459$ , found identically by unordered and by ordered counting.

Use the complement for an "at least one" card event

Five cards are dealt from a standard pack of $52$ . Find the probability that the hand contains at least one ace.

Identify the sample space. A hand is an unordered selection of $5$ from $52$ :

n(S) = {}^{52}C_{5} = 2\,598\,960.

Switch to the complement. "At least one ace" splits into exactly one, two, three or four aces, so instead count the complement, a hand with no ace, chosen from the $48$ non-aces:

n(\text{no ace}) = {}^{48}C_{5} = 1\,712\,304, \qquad P(\text{no ace}) = \frac{1\,712\,304}{2\,598\,960} = \frac{35\,673}{54\,145}.

Subtract from $1$ .

P(\text{at least one ace}) = 1 - \frac{35\,673}{54\,145} = \frac{54\,145 - 35\,673}{54\,145} = \frac{18\,472}{54\,145} \approx 0.3412.

Why the complement wins. Counting the four "exactly" cases would need $^{4}C_{1}\,^{48}C_{4} + {}^{4}C_{2}\,^{48}C_{3} + {}^{4}C_{3}\,^{48}C_{2} + {}^{4}C_{4}\,^{48}C_{1}$ ; the single complement count replaces all four.

Final answer. $P(\text{at least one ace}) = \dfrac{18\,472}{54\,145} \approx 0.3412$ .

Find five-digit-number probabilities, including inclusion-exclusion

A five-digit number is chosen at random (the first digit is not $0$ ). Find the probability that (a) the number contains at least one $4$ , and (b) the number contains at least one $4$ and at least one $5$ .

Count the sample space. The first digit has $9$ choices and the other four have $10$ each:

n(S) = 9 \times 10^4 = 90\,000.

Part (a): complement of "at least one $4$ ". A number with no $4$ has $8$ choices for the first digit ( $1$ to $9$ , not $4$ ) and $9$ for each of the others ( $0$ to $9$ , not $4$ ):

n(\text{no } 4) = 8 \times 9^4 = 52\,488, \qquad P(\ge 1 \text{ four}) = \frac{90\,000 - 52\,488}{90\,000} = \frac{37\,512}{90\,000} = \frac{521}{1250} \approx 0.4168.

Part (b): the complement is an "or", so use inclusion-exclusion. Let $A$ be "no $4$ " and $B$ be "no $5$ ". The event "at least one $4$ and at least one $5$ " fails exactly when there is no $4$ or no $5$ , that is on $A \cup B$ . With $n(A) = n(B) = 8 \times 9^4 = 52\,488$ and $n(A \cap B) = 7 \times 8^4 = 28\,672$ (no $4$ and no $5$ ),

n(A \cup B) = 52\,488 + 52\,488 - 28\,672 = 76\,304.

Subtract from the sample space.

P(\ge 1 \text{ four and} \ge 1 \text{ five}) = \frac{90\,000 - 76\,304}{90\,000} = \frac{13\,696}{90\,000} = \frac{856}{5625} \approx 0.1522.

Final answer. (a) $\dfrac{521}{1250} \approx 0.4168$ ; (b) $\dfrac{856}{5625} \approx 0.1522$ .

Compare with and without replacement

A bag holds $3$ red, $7$ yellow and $5$ blue discs, $15$ in all. Three discs are drawn. Find the probability that (a) all three are yellow when drawn together (without replacement), (b) all three are the same colour without replacement, and (c) all three are yellow when each disc is replaced before the next draw.

Part (a): without replacement is a ratio of combinations. The sample space is $^{15}C_{3} = 455$ , and the all-yellow selections number $^{7}C_{3} = 35$ :

P(\text{all yellow}) = \frac{^{7}C_{3}}{^{15}C_{3}} = \frac{35}{455} = \frac{1}{13} \approx 0.0769.

Part (b): all one colour, by disjoint cases. All red, all yellow or all blue are mutually exclusive, so add the favourable counts $^{3}C_{3} + {}^{7}C_{3} + {}^{5}C_{3} = 1 + 35 + 10 = 46$ :

P(\text{all same colour}) = \frac{46}{455} \approx 0.1011.

Part (c): with replacement, multiply independent draws. Each draw is yellow with probability $\dfrac{7}{15}$ , and the three draws are independent:

P(\text{all yellow, with replacement}) = \left(\frac{7}{15}\right)^3 = \frac{343}{3375} \approx 0.1016.

Compare (a) and (c). With replacement keeps the yellow share at $\dfrac{7}{15}$ every draw, so it is slightly more likely than the without-replacement value $\dfrac{1}{13} \approx 0.0769$ , where each yellow taken leaves fewer behind.

Final answer. (a) $\dfrac{1}{13} \approx 0.0769$ ; (b) $\dfrac{46}{455} \approx 0.1011$ ; (c) $\dfrac{343}{3375} \approx 0.1016$ .

Common traps

Mixing ordered and unordered between the two counts: If the sample space is unordered ( $^{n}C_{r}$ ), the favourable count must be unordered too; if it is ordered ( $^{n}P_{r}$ or a power), so must the favourable count be. Counting one part ordered and the other unordered gives a wrong ratio. Decide once, then stick to it.
Adding when you should multiply, or vice versa: Choosing fixed numbers from separate pools (men and women, hearts and non-hearts) multiplies the combinations. Splitting into mutually exclusive cases (exactly $2$ , then $3$ , then $4$ ) adds the counts. Multiply within one selection, add across alternative cases.
Counting "at least one" directly: "At least one ace" or "at least one $4$ " almost always has a shorter complement, "no ace" or "no $4$ ". Counting the cases one by one invites a missed case; $1 - P(\text{none})$ does not.
Forgetting the overlap in an "or": $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ . Two "at least one" conditions joined by "and" complement to an "or" that overlaps, so the overlap $A \cap B$ must be subtracted exactly once.
Mishandling the leading-digit restriction: A five-digit number cannot start with $0$ , so the first slot has fewer choices than the others ( $9$ , not $10$ ), and a "no $4$ " first slot has $8$ , not $9$ . Apply the restriction to the first slot only.
Confusing with and without replacement: "Drawn together" or "all at once" means without replacement, a ratio of combinations. "Replaced before the next draw" means independent draws whose probabilities multiply. The two give different answers, so read the wording carefully.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksA club has

9

members, of whom

5

are men and

4

are women. A committee of

3

is chosen at random. (i) How many different committees are possible? (ii) Find the probability that the committee consists entirely of men.

Show worked solution →

Part (i): the total is an unordered selection. A committee is unordered, so count combinations. Choose any $3$ of the $9$ members:

^{9}C_{3} = \frac{9 \times 8 \times 7}{3!} = \frac{504}{6} = 84.

Part (ii): the favourable count is the all-men committees. All $3$ members come from the $5$ men:

^{5}C_{3} = \frac{5 \times 4 \times 3}{3!} = 10.

Form the ratio favourable over total. Both counts use the same equally likely committees, so

P(\text{all men}) = \frac{^{5}C_{3}}{^{9}C_{3}} = \frac{10}{84} = \frac{5}{42} \approx 0.119.

Answer. (i) $84$ committees; (ii) $P(\text{all men}) = \dfrac{5}{42} \approx 0.119$ .

foundation3 marksThe integers

1

10

are written on ten cards. Four cards are drawn at random (all at once). Find the probability that (i) the card numbered

9

is among those drawn, and (ii) the card numbered

8

is not drawn.

Show worked solution →

Count the total selections. Four cards chosen from ten, order irrelevant:

^{10}C_{4} = \frac{10 \times 9 \times 8 \times 7}{4!} = \frac{5040}{24} = 210.

Part (i): force card $9$ in, then choose the rest. With $9$ already drawn, the other $3$ cards come from the remaining $9$ , giving $^{9}C_{3} = 84$ favourable selections:

P(9 \text{ drawn}) = \frac{^{9}C_{3}}{^{10}C_{4}} = \frac{84}{210} = \frac{2}{5} = 0.4.

Part (ii): ban card $8$ , then choose freely. Removing $8$ leaves $9$ cards, from which all $4$ are drawn, giving $^{9}C_{4} = 126$ favourable selections:

P(8 \text{ not drawn}) = \frac{^{9}C_{4}}{^{10}C_{4}} = \frac{126}{210} = \frac{3}{5} = 0.6.

Answer. (i) $\dfrac{2}{5} = 0.4$ ; (ii) $\dfrac{3}{5} = 0.6$ .

core4 marksFive cards are dealt from a standard pack of

52

. Find the probability that (i) all five cards are hearts, and (ii) exactly one of the five cards is a heart. Leave each answer as a fraction and as a decimal to four places.

Show worked solution →

Count the total number of hands. A hand is an unordered selection of $5$ from $52$ :

^{52}C_{5} = 2\,598\,960.

Part (i): all five from the $13$ hearts. The favourable count is $^{13}C_{5} = 1287$ , so

P(\text{all hearts}) = \frac{^{13}C_{5}}{^{52}C_{5}} = \frac{1287}{2\,598\,960} = \frac{33}{66\,640} \approx 0.0005.

Part (ii): one heart and four non-hearts, multiplied. Choose $1$ of the $13$ hearts and $4$ of the $39$ non-hearts, then multiply (separate pools):

^{13}C_{1} \times {}^{39}C_{4} = 13 \times 82\,251 = 1\,069\,263.

Hence

P(\text{exactly one heart}) = \frac{1\,069\,263}{2\,598\,960} = \frac{27\,417}{66\,640} \approx 0.4114.

Answer. (i) $\dfrac{33}{66\,640} \approx 0.0005$ ; (ii) $\dfrac{27\,417}{66\,640} \approx 0.4114$ .

core3 marksA five-digit number is chosen at random (so the first digit is not

0

). Find the probability that all five of its digits are different. Leave the answer as a fraction and as a decimal to four places.

Show worked solution →

Count the total of five-digit numbers (the sample space). The first digit has $9$ choices ( $1$ to $9$ ), and each of the other four has $10$ choices:

9 \times 10^4 = 90\,000.

Count the favourable numbers slot by slot, without repetition. The first digit avoids $0$ , so $9$ choices; the second avoids only the first used digit, so $9$ choices (the $0$ is now back in play); then $8$ , then $7$ , then $6$ :

9 \times 9 \times 8 \times 7 \times 6 = 27\,216.

Form the ratio.

P(\text{all digits different}) = \frac{27\,216}{90\,000} = \frac{189}{625} = 0.3024.

Why the second slot still has $9$ . The first digit removed one value but freed nothing; the digit $0$ , banned only from the first slot, becomes available from the second slot onward, so the falling pattern is $9, 9, 8, 7, 6$ rather than $9, 8, 7, 6, 5$ .

Answer. $\dfrac{189}{625} = 0.3024$ .

exam4 marksA committee of

5

is chosen at random from

6

Labor and

5

Liberal members of a council. Find the probability that Labor holds a majority on the committee (that is, at least

3

of the

5

members are from Labor). Leave the answer as a fraction and as a decimal to four places.

Show worked solution →

Count the total committees. Five chosen from $11$ , order irrelevant:

^{11}C_{5} = 462.

Read 'majority' as the cases $3$ , $4$ or $5$ Labor, and add them. Each case fixes how many come from each pool, so multiply within a case and add across cases. The favourable count is

^{6}C_{3}\,^{5}C_{2} + {}^{6}C_{4}\,^{5}C_{1} + {}^{6}C_{5}\,^{5}C_{0}.

Evaluate each case.

^{6}C_{3}\,^{5}C_{2} = 20 \times 10 = 200, \quad ^{6}C_{4}\,^{5}C_{1} = 15 \times 5 = 75, \quad ^{6}C_{5}\,^{5}C_{0} = 6 \times 1 = 6.

Add and form the ratio. The favourable total is $200 + 75 + 6 = 281$ , so

P(\text{Labor majority}) = \frac{281}{462} \approx 0.6082.

Answer. $\dfrac{281}{462} \approx 0.6082$ .

exam4 marksA bag holds

7

white and

5

black discs. Three discs are taken. Find the probability that all three are black if (i) the discs are taken without replacement (all at once), and (ii) the discs are taken one at a time, each replaced before the next is drawn. Leave each answer as a fraction and as a decimal to four places.

Show worked solution →

Part (i): without replacement is an unordered selection. The total is $^{12}C_{3} = 220$ and the favourable count is the all-black selections $^{5}C_{3} = 10$ :

P(\text{3 black, no replacement}) = \frac{^{5}C_{3}}{^{12}C_{3}} = \frac{10}{220} = \frac{1}{22} \approx 0.0455.

Part (ii): with replacement, each draw is independent. The bag is restored each time, so each draw is black with probability $\dfrac{5}{12}$ , and the three draws multiply:

P(\text{3 black, with replacement}) = \left(\frac{5}{12}\right)^3 = \frac{125}{1728} \approx 0.0723.

Compare the two. Replacement keeps $5$ black among $12$ every time, so it is slightly more likely to draw three blacks than the without-replacement case, where each black removed leaves fewer for the next draw.

Answer. (i) $\dfrac{1}{22} \approx 0.0455$ ; (ii) $\dfrac{125}{1728} \approx 0.0723$ .

What this dot point is asking

The answer

The equally likely model: probability as a ratio of two counts

Ordered or unordered? Choose once, and use it for both counts

Selection probabilities with combinations

Exactly kkk of a type, and the separate-pools rule

Card-hand probabilities

"At least one" by the complement

A five-digit-number probability, and inclusion-exclusion

With replacement versus without replacement

How exam questions ask about this dot point

Practice questions

Related dot points

Exactly $k$ of a type, and the separate-pools rule