Year 9 Maths Probability and Statistics

Using Arrays for Two-Step Experiments

Q: When can you multiply probabilities?

To find P(A and B), multiply P(A) by P(B) when the events are independent. For dependent events, multiply by the probability of B after A has happened.

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Theory

A two-step experiment has two parts, such as tossing a coin and rolling a die. This Year 9 maths page shows how to list every outcome in an array (grid), count the outcomes for an event, multiply probabilities for independent events, and handle drawing with and without replacement. It also uses the probabilities to predict expected results.

A two-step (multistage) experiment has two parts, such as rolling two dice or drawing two cards. Each outcome is a pair, such as \((3, 5)\).

The number of outcomes is the number for step one times the number for step two: a coin and a die give \(2 \times 6 = 12\) outcomes.

Events are independent if the result of one does not change the probability of the other, such as two separate dice. They are dependent if it does, such as drawing a second card without replacement.

The expected number of results in \(n\) trials is \(P(\text{event}) \times n\).

The relative frequency (observed probability) from a simulation or experiment is the number of times the event happened divided by the number of trials.

A grid (array) lists every outcome of a two-step experiment. Count the cells for the event and divide by the total.

Fill each cell with the sum. A sum of \(5\) is in \(4\) of the \(36\) cells, so \(P(\text{sum } 5) = \dfrac{4}{36} = \dfrac{1}{9}\).

For independent events \(A\) and \(B\), and for repeated trials:

\[P(A \text{ and } B) = P(A) \times P(B)\]

\[\text{expected number} = P(\text{event}) \times \text{trials}\]

P (A and B) = P (A) \times P (B)

\[\text{relative frequency} = \dfrac{\text{times it happened}}{\text{number of trials}}\]

Grids for two dice. Put one die down the side and one across the top. Fill each cell with the pair, the sum, the difference or the product, then count the cells you need.

Arrays without replacement. When the same item cannot be picked twice, cross out the cells on the diagonal (the same card both times). Four cards give \(4 \times 4 - 4 = 12\) outcomes, not \(16\).

Two orders. "One of each" can happen two ways. With \(3\) red and \(2\) green, drawing two without replacement: \(P = \dfrac{3}{5} \times \dfrac{2}{4} + \dfrac{2}{5} \times \dfrac{3}{4} = \dfrac{12}{20} = \dfrac{3}{5}\).

Using the complement. A product of two numbers is odd only if both are odd, so \(P(\text{even product}) = 1 - P(\text{both odd})\).

Estimating from a simulation. If an event happened in \(28\) of \(40\) simulated days, estimate \(\dfrac{28}{40} \times 365 \approx 256\) days in a year.

How to find a probability in a two-step experiment

List the outcomes in a grid (or a tree) and count them.
Mark the outcomes in the event, counting each pair once.
Divide by the total number of outcomes.

How to use the multiplication rule

Check whether the steps are independent. If not (no replacement), change the second probability: one fewer item in total, and one fewer of the type taken.
Multiply the probabilities of the steps.
Add the results if the event can happen in more than one order.

Example 1 — Using a 6 by 6 grid

Two fair six-sided dice are rolled. Find the probability that the larger of the two numbers is \(4\) (a double \(4\) counts).

Solution

There are \(6 \times 6 = 36\) equally likely outcomes. List the pairs whose largest number is \(4\):

\(\text{1st} = 4\)	\(:\)	\((4,1), (4,2), (4,3), (4,4)\)
\(\text{2nd} = 4\)	\(:\)	\((1,4), (2,4), (3,4)\)

That is \(7\) outcomes (\((4,4)\) is listed only once).

\(P\)

\(=\)

\(\boldsymbol{\dfrac{7}{36}}\)

P = \frac{7}{36}

Example 2 — Independent events

The probability that Ava's bus is on time is \(0.85\). The probability that her train is on time is \(0.7\). The two are independent. Find the probability that:
(i) both are on time
(ii) both are late.

Solution

Independent events: multiply.

(i) Both on time:

\(P\)	\(=\)	\(0.85 \times 0.7\)
\(\)	\(=\)	\(\boldsymbol{0.595}\)

P = 0.85 \times 0.7 = 0.595

(ii) First find each probability of being late with the complement.

\(P(\text{bus late})\)	\(=\)	\(1 - 0.85 = 0.15\)
\(P(\text{train late})\)	\(=\)	\(1 - 0.7 = 0.3\)
\(P(\text{both late})\)	\(=\)	\(0.15 \times 0.3\)
\(\)	\(=\)	\(\boldsymbol{0.045}\)

P = 0.15 \times 0.3 = 0.045

Example 3 — With and without replacement

Six cards are numbered \(1\) to \(6\). Two cards are drawn at random. Find the probability that both are odd if the first card is:
(i) put back before the second draw
(ii) not put back.

Solution

The odd cards are \(1\), \(3\) and \(5\).

(i) With replacement, the second draw is the same as the first.

\(P\)

\(=\)

\(\dfrac{3}{6} \times \dfrac{3}{6} = \dfrac{9}{36} = \boldsymbol{\dfrac{1}{4}}\)

(ii) Without replacement, one odd card has gone: \(2\) odd cards out of \(5\) are left.

\(P\)

\(=\)

\(\dfrac{3}{6} \times \dfrac{2}{5} = \dfrac{6}{30} = \boldsymbol{\dfrac{1}{5}}\)

P = \frac{1}{5}

The second events are dependent in (ii): the first draw changes the second.

Example 4 — Expected results and a simulation

In a game, two dice are rolled and a player wins if the total is \(7\).
(i) Find the probability of winning.
(ii) How many wins would you expect in \(240\) games?
(iii) A computer simulation of \(300\) games gave \(46\) wins. Find the relative frequency and compare.

Solution

(i) The totals of \(7\) are \((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\): \(6\) of \(36\) outcomes.

\(P(\text{win})\)	\(=\)	\(\dfrac{6}{36}\)
\(\)	\(=\)	\(\boldsymbol{\dfrac{1}{6}}\)

P = \frac{6}{36} = \frac{1}{6}

(ii) Multiply the probability by the number of games.

\(\text{expected}\)	\(=\)	\(\dfrac{1}{6} \times 240\)
\(\)	\(=\)	\(\boldsymbol{40}\)

\frac{1}{6} \times 240 = 40

(iii) Divide the wins by the number of games played.

\(\text{rel. freq.}\)	\(=\)	\(\dfrac{46}{300}\)
\(\)	\(\approx\)	\(\boldsymbol{0.153}\)

\frac{46}{300} = 0.153

This is close to \(\dfrac{1}{6} \approx 0.167\). More trials would usually bring it closer.

Common pitfalls

Order matters in a grid. \((2, 5)\) and \((5, 2)\) are different outcomes with two dice. A double such as \((3, 3)\) appears once.

Change both numbers without replacement. After taking \(1\) of \(3\) odd cards from \(6\), the next is \(\dfrac{2}{5}\), not \(\dfrac{3}{6}\) or \(\dfrac{2}{6}\).

Multiply, do not add, for "and". Both on time is \(0.85 \times 0.7\), not \(0.85 + 0.7\) (that is more than \(1\)).

Expected is not guaranteed. With \(P = \dfrac{1}{6}\), \(60\) rolls should give about \(10\) results, but not exactly \(10\) every time.

Frequently asked questions

How many outcomes does a two-step experiment have?

Multiply the number of outcomes of each step. Two six-sided dice have \(6 \times 6 = 36\) outcomes; a coin and a die have \(12\).

What is the difference between independent and dependent events?

Independent events do not affect each other, like two coin tosses. Dependent events do, like drawing a second card after the first is kept out.

When can you multiply probabilities?

To find \(P(A \text{ and } B)\), multiply \(P(A)\) by \(P(B)\) when the events are independent. For dependent events, multiply by the probability of \(B\) after \(A\) has happened.

What is relative frequency?

It is the fraction of trials in which an event actually happened: the count divided by the number of trials. It estimates the probability.

How do you find the expected number of results?

Multiply the probability by the number of trials. For a probability of \(\dfrac{1}{4}\) over \(120\) games, expect about \(30\) results.

Using Arrays for Two-Step Experiments

📖 Theory

How to find a probability in a two-step experiment

How to use the multiplication rule

Common pitfalls

Frequently asked questions

Theory