Year 10 Maths - 10A Probability

Three-Step Experiments, With and Without Replacement

Q: How do you find the probability of at least one success in n trials?

Find the probability of no successes, which is (1 minus p) to the power n, then subtract it from 1.

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Theory

A three-step experiment repeats a chance step three (or more) times. With replacement every stage is the same and the trials are independent; without replacement the total and the count of the item taken both drop by \(1\) at each stage. This Year 10A maths page covers at least one in \(n\) trials, exactly \(k\) successes by counting orders, the first success, three draws from a bag, cards and raffles, and working backwards. In the Australian Curriculum three-step experiments are Year 9 content, so there it is revision. Curriculum: VC2M10P02 / AC9M9P01 (Year 9, revised).

A three-step experiment has three stages, such as three draws or three throws. Its outcomes are lists such as \(RRB\), and the order matters. Experiments with four or more stages work the same way.

Sampling with replacement means each chosen item is returned before the next choice. The bag is the same every time, so each draw is independent of the others.

The same model covers any repeated independent trials with a fixed chance of success \(p\): rolling a die, tossing a coin, shooting at a target, digits of a code chosen with repeats allowed.

Sampling without replacement means an item, once chosen, is not put back. Choosing a committee, dealing cards, drawing raffle prizes and taking items from a box to test are all without replacement.

The draws are dependent: each probability after the first is a conditional probability, worked out from what is left. If the first item was red, the second draw has one fewer red and one fewer item in total.

Drawing several items "at once" or "at random together" gives the same probabilities as drawing them one after another without replacement, so use the same method.

Multiply along a path: \(n\) successes in a row have probability \(p^n\), and \(n\) failures have \((1-p)^n\). An event that can happen in several orders, such as "exactly two out of three", is a sum of equal paths: number of orders \(\times\) probability of one order.

"At least one" is the complement of "none": \(P(\text{at least one})=1-(1-p)^n\). As \(n\) grows this gets closer to \(1\), but never reaches it.

With replacement every stage is identical, so each path with the same number of successes has the same probability. Count the paths, then multiply.

Without replacement both the count of the colour drawn and the total go down by \(1\) at each step (Example 2).

With replacement (independent trials, chance \(p\) each time):

\[P(\text{all } n)=p^n\]

\[P(\text{none})=(1-p)^n\]

\[P(\text{at least one})=1-(1-p)^n\]

P (at least one) = 1 - {(1 - p)}^{n}

Trials	Successes wanted	Number of orders (paths)
\(2\)	exactly \(1\)	\(2\): \(SF\), \(FS\)
\(3\)	exactly \(1\) or exactly \(2\)	\(3\)
\(4\)	exactly \(1\) or exactly \(3\)	\(4\)
\(4\)	exactly \(2\)	\(6\): \(SSFF\), \(SFSF\), \(SFFS\), \(FSSF\), \(FSFS\), \(FFSS\)
\(n\)	exactly \(1\) or exactly \(n-1\)	\(n\)
\(3\)	three different things	\(3\times 2\times 1=6\)

No repeats allowed. If a \(3\)-symbol code uses the digits \(1\) to \(5\) with repeats allowed, the chance that all three are different is \(\dfrac{5}{5}\times\dfrac{4}{5}\times\dfrac{3}{5}=\dfrac{12}{25}\): each new symbol must avoid the ones already used.

Without replacement, with \(a\) of type \(A\) and \(b\) of type \(B\) out of \(N\):

\[P(AA)=\dfrac{a}{N}\times\dfrac{a-1}{N-1}\]

\[P(AB)=\dfrac{a}{N}\times\dfrac{b}{N-1}\]

P = \frac{a}{N} \times \frac{a - 1}{N - 1}

	With replacement	Without replacement
second-draw total	same as the first	one less
draws are	independent	dependent
two of a kind, \(a\) of \(N\)	\(\left(\dfrac{a}{N}\right)^2\)	\(\dfrac{a}{N}\times\dfrac{a-1}{N-1}\) (smaller)
at least one of a kind	\(1-\left(\dfrac{N-a}{N}\right)^2\)	\(1-\dfrac{N-a}{N}\times\dfrac{N-a-1}{N-1}\) (larger)

Raffles. If you hold \(t\) of \(N\) tickets and \(k\) prizes are drawn, it is quickest to find \(P(\text{no prize})=\dfrac{N-t}{N}\times\dfrac{N-t-1}{N-1}\times\cdots\) (\(k\) factors) and subtract from \(1\). Choosing your own numbers in a draw works the same way: the first winning number must be one of yours, then the next one of your remaining numbers.

Any position is equally likely. When items are taken out one by one until none are left, every item is equally likely to be in any position. So the chance that the item in a given position is a particular type equals the fraction of that type at the start.

How to solve a with-replacement problem

Find the probability of each outcome for one draw or trial. It stays the same every time.
Describe the event as paths: "all", "none", "exactly \(k\)", "first on trial \(k\)".
Multiply along one path. If several paths have the same mix of outcomes, multiply by the number of orders.
Use the complement for "at least one" and for "not all".
Round only at the end, using exact values or the calculator memory.

How to solve a without-replacement problem

Write down the numbers of each type and the total.
Follow one path: for each draw, the numerator is how many of that type are left, and the denominator is how many items are left.
Multiply along the path. For "exactly \(k\)", multiply by the number of orders.
Add different cases, such as "same colour" \(=RR+BB\), or "more red than blue" \(=RRR+\) two red.
Use the complement for "at least one" and "not all". "At least one from each group" is \(1-P(\text{all from one group})\) summed over the groups.

Other question types: without replacement

Letters and labels: count repeated letters separately. A word with \(3\) A tiles has \(3\) chances of drawing an A.
Unknown count: write the path with \(n\) and solve. A product of consecutive whole numbers, such as \(r(r-1)(r-2)=120\), can be solved by recognising \(4\times 5\times 6\).
Comparing methods: for the same box, sampling without replacement makes "two of the same rare type" less likely and "at least one of a type" more likely than with replacement.
Show-that algebra: with \(n\) of each of two colours, write both same-colour paths over \(2n\) and \(2n-1\), then simplify.

Example 1 — At least one in n trials

(i) A basketballer makes each free throw with probability \(0.8\), independently. In \(5\) throws, find \(P(\text{all } 5)\) and \(P(\text{misses at least one})\), to \(4\) decimal places.
(ii) A game square gives a prize with probability \(0.15\) on each turn. Find the smallest number of turns for which \(P(\text{at least one prize})\gt 0.75\).

Solution

(i) Independent trials, so multiply.

\(P(\text{all } 5)\)	\(=\)	\(0.8^5=0.32768\)
\(\)	\(\approx\)	\(\boldsymbol{0.3277}\)

Missing at least one is the complement of making all \(5\).

\(P(\text{a miss})\)	\(=\)	\(1-0.8^5\)
\(\)	\(\approx\)	\(\boldsymbol{0.6723}\)

(ii) \(P(\text{at least one})=1-0.85^n\). This is more than \(0.75\) when \(0.85^n\lt 0.25\). Try values:

\(0.85^8\)	\(=\)	\(0.272\ldots \quad (\text{too big})\)
\(0.85^9\)	\(=\)	\(0.231\ldots \quad (\lt 0.25)\)

The smallest number is \(\boldsymbol{n=9}\) turns.

Example 2 — Exactly k successes

A biased coin shows a head with probability \(0.6\). It is tossed \(4\) times. Find:
(i) \(P(\text{exactly } 3 \text{ heads})\)
(ii) \(P(\text{the first head is on the fourth toss})\)

Solution

(i) The single tail can be in any of the \(4\) positions: \(THHH\), \(HTHH\), \(HHTH\), \(HHHT\). Each path has three \(0.6\)s and one \(0.4\).

\(P\)	\(=\)	\(4\times 0.6^3\times 0.4\)
\(\)	\(=\)	\(4\times 0.0864=\boldsymbol{0.3456}\)

(ii) The first head on toss four is one path only: \(TTTH\).

\(P\)

\(=\)

\(0.4^3\times 0.6=\boldsymbol{0.0384}\)

P = 4 \times {0.6}^{3} \times 0.4

Example 3 — Three draws

A bag has \(6\) red and \(4\) blue marbles. Three are drawn at random without replacement. Find:
(i) \(P(\text{all red})\)
(ii) \(P(\text{exactly two red})\)
(iii) \(P(\text{at least one blue})\)

Solution

(i) The totals go \(10\), \(9\), \(8\).

\(P(RRR)\)

\(=\)

\(\dfrac{6}{10}\times\dfrac{5}{9}\times\dfrac{4}{8}=\dfrac{120}{720}=\boldsymbol{\dfrac{1}{6}}\)

(ii) Exactly two red: \(RRB\), \(RBR\), \(BRR\). Each order has numerators \(6, 5, 4\) in some order over \(10\times 9\times 8\), so each equals \(\dfrac{120}{720}\).

\(P\)

\(=\)

\(3\times\dfrac{120}{720}=\boldsymbol{\dfrac{1}{2}}\)

(iii) At least one blue is everything except \(RRR\): \(1-\dfrac{1}{6}=\boldsymbol{\dfrac{5}{6}}\).

P (R R R) = \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}

Example 4 — Finding the number of items

A box has \(n\) green and \(2\) red counters. Two counters are drawn without replacement, and \(P(\text{both green})=\dfrac{2}{5}\).
(i) Find \(n\).
(ii) Compare with the probability of two greens if the first counter is replaced.

Solution

(i) There are \(n+2\) counters, then \(n+1\).

\(\dfrac{n}{n+2}\times\dfrac{n-1}{n+1}\)	\(=\)	\(\dfrac{2}{5}\)
\(5n(n-1)\)	\(=\)	\(2(n+2)(n+1)\)
\(5n^2-5n\)	\(=\)	\(2n^2+6n+4\)
\(3n^2-11n-4\)	\(=\)	\(0\)
\((3n+1)(n-4)\)	\(=\)	\(0\)

\(n\) is a positive whole number, so \(\boldsymbol{n=4}\). Check: \(\dfrac{4}{6}\times\dfrac{3}{5}=\dfrac{2}{5}\).

(ii) With replacement: \(\left(\dfrac{4}{6}\right)^2=\dfrac{4}{9}\), which is more than \(\dfrac{2}{5}\). Removing a green counter makes a second green less likely.

Common pitfalls

At least one is not n × p. Three tries at \(0.3\) do not give \(0.9\). The correct value is \(1-0.7^3=0.657\).

"First success on trial 3" is one path. It is fail, fail, success only, not \(3\) orders.

Do not round early. Rounding \(0.85^8\) to \(0.27\) then subtracting can change a 4-decimal-place answer.

Reduce the total every time. Three draws from \(10\) items use denominators \(10\), \(9\), \(8\), not \(10\) each time.

Only reduce the type that was drawn. After a king is drawn, there are still \(4\) queens left, out of \(51\) cards.

Count every order. "One red and one blue" is \(RB\) or \(BR\). Leaving one out halves the answer.

Check unknown counts are whole numbers. A quadratic for \(n\) may give a negative or fractional root. Only a positive whole number can be a number of items.

Frequently asked questions

What does sampling with replacement mean?

Each item chosen is put back before the next choice, so the population is the same every time. The draws are independent and each has the same probabilities.

What does sampling without replacement mean?

Each item chosen is kept out, so it cannot be chosen again. The total number of items, and the number of the type chosen, both go down by \(1\) before the next draw.

How do you find the probability of at least one success in n trials?

Find the probability of no successes, \((1-p)^n\), then subtract it from \(1\).

How do you find the probability of exactly k successes?

Work out the probability of one order, such as success, success, failure, then multiply by the number of different orders. For exactly \(2\) successes in \(4\) trials there are \(6\) orders.

Is drawing two items at once the same as drawing one after the other?

Yes. Taking two items together gives the same probabilities as taking one and then another without putting the first back.

Are draws without replacement independent?

No. Each draw changes what is left, so the probabilities for the next draw depend on what was drawn before. They are dependent events.

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