Year 10 Maths - 10 Probability

Review of Probability

Q: How do you write a probability as a percentage?

Multiply the fraction or decimal by 100. For example 3 out of 8 is 0.375, which is 37.5 percent.

Q: How do you find the number of items from a probability?

Use the probability as parts of the whole. If 5 out of 8 parts are white and there are 25 white items, one part is 5 items and the whole is 40.

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Theory

Probability measures how likely an event is, from \(0\) (impossible) to \(1\) (certain). For equally likely outcomes it is the number of favourable outcomes divided by the total in the sample space. This Year 10 maths page revises the probability from Years 7 to 9: listing outcomes, fractions, decimals and percentages, complements, ratios, expected numbers, and working backwards to a missing number of items.

A chance experiment is something with an uncertain result, such as rolling a die. Each possible result is an outcome.

The sample space \(S\) is the list of all possible outcomes. For one roll of a die, \(S = \{1, 2, 3, 4, 5, 6\}\).

An event is a set of outcomes, such as "rolling an even number". The outcomes in the event are the favourable outcomes.

Outcomes are equally likely when each has the same chance, as with a fair die, a spinner with equal sectors, or choosing "at random".

The probability \(P(\text{event})\) is a number from \(0\) to \(1\). It can be written as a fraction, a decimal or a percentage.

The complement of an event is everything that is not in the event, such as "not blue". An event and its complement have probabilities that add to \(1\).

The sectors are equal, so each of the \(8\) outcomes is equally likely. Three are red: \(P(\text{red}) = \dfrac{3}{8}\).

List the sample space, then count the favourable outcomes: \(P = \dfrac{2}{6} = \dfrac{1}{3}\).

For equally likely outcomes:

\[P(\text{event}) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}\]

P (event) = \frac{favourable outcomes}{total outcomes}

Repeated letters. Count every card, including repeats. In BANANA there are \(6\) cards and \(3\) of them are A, so \(P(\text{A}) = \dfrac{3}{6} = \dfrac{1}{2}\).

Different letters. "A letter of the alphabet that is in the word PARROT" counts each letter once: P, A, R, O, T is \(5\) different letters, so \(P = \dfrac{5}{26}\). Repeats only count when every letter is on its own card.

Counting whole numbers. From \(25\) to \(60\) there are \(60 - 25 + 1 = 36\) numbers. Remember that \(1\) is not prime: the primes up to \(20\) are \(2, 3, 5, 7, 11, 13, 17, 19\).

Ratios. Red to blue \(2 : 7\) means \(9\) parts, so \(P(\text{red}) = \dfrac{2}{9}\).

Things that change. If an item is taken out and not put back, the total goes down by one. After one of \(300\) raffle tickets is drawn, a second draw is from \(299\) tickets. When only one position decides the event (such as the first digit of a number made from shuffled digits), just look at that position.

All the probabilities add to 1. If a spinner shows only red, blue and other colours with \(P(\text{red}) = 0.3\) and \(P(\text{blue}) = 0.45\), then \(P(\text{other}) = 1 - 0.3 - 0.45 = 0.25\). With an unknown, write the sum as an equation: if \(P(\text{red}) = 0.25\) and \(P(\text{green})\) is \(0.15\) more than \(P(\text{yellow}) = y\), then \(0.25 + y + (y + 0.15) = 1\), so \(2y = 0.6\) and \(y = 0.3\).

Expected number of times an event happens in a number of trials:

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

expected number = P (event) \times n

where \(n\) is the number of trials.

Example. A fair die is rolled \(300\) times. A number greater than \(4\) is expected \(\dfrac{2}{6} \times 300 = 100\) times.

How to find a probability

Check that the outcomes are equally likely.
Count the total number of outcomes (list them if it helps).
Count the favourable outcomes. For "not" or "but not", cross out the ones that do not fit.
Divide and simplify. Change to a decimal or percentage if asked.

How to work backwards

Find the probability of the group you know the number of (use \(1 -\) the given probability if needed).
Use parts: the fraction tells you how many parts that group is; divide to find one part.
Or write an equation such as \(\dfrac{5 + k}{20 + k} = \dfrac{2}{5}\) and solve it. Check your answer.

Complements and expected numbers

"Not" an event: work out the event, then use \(P(\text{not } A) = 1 - P(A)\).
A missing probability: add the probabilities you know and subtract from \(1\). If two of them are linked, use a letter and solve.
Expected number: multiply the probability by the number of trials. It is a long-run average, not a promise.

Example 1 — Counting favourable outcomes

The numbers \(1\) to \(40\) are written on cards and one card is chosen at random. Find the probability that the number is:
(i) a multiple of \(6\)
(ii) a multiple of \(5\) but NOT a multiple of \(10\).

Solution

There are \(40\) equally likely outcomes.

(i) Multiples of \(6\): \(6, 12, 18, 24, 30, 36\). That is \(6\) cards.

\(P(\text{multiple of } 6)\)

\(=\)

\(\dfrac{6}{40} = \boldsymbol{\dfrac{3}{20}}\)

(ii) Multiples of \(5\): \(5, 10, 15, 20, 25, 30, 35, 40\). Cross out the multiples of \(10\), leaving \(5, 15, 25, 35\).

\(P\)

\(=\)

\(\dfrac{4}{40} = \boldsymbol{\dfrac{1}{10}}\)

P = \frac{1}{10}

Example 2 — Fraction, decimal and percentage

A bag holds \(6\) red, \(5\) green and \(9\) white counters. One counter is drawn at random. Find:
(i) \(P(\text{green})\)
(ii) \(P(\text{not white})\). Give each as a fraction, a decimal and a percentage.

Solution

Total: \(6 + 5 + 9 = 20\) counters.

\(P(\text{green})\)	\(=\)	\(\dfrac{5}{20} = \dfrac{1}{4}\)
\(\)	\(=\)	\(\boldsymbol{0.25 = 25\%}\)

(ii) "Not white" means red or green: \(6 + 5 = 11\) counters.

\(P(\text{not white})\)	\(=\)	\(\dfrac{11}{20}\)
\(\)	\(=\)	\(\boldsymbol{0.55 = 55\%}\)

P = \frac{11}{20} = 0.55

Example 3 — Working backwards to a total

A bag holds only blue and white beads. The probability of drawing a blue bead is \(\dfrac{3}{8}\). There are \(25\) white beads. How many beads are in the bag, and how many are blue?

Solution

The beads are blue or white, so \(P(\text{white}) = 1 - \dfrac{3}{8} = \dfrac{5}{8}\).

So \(5\) parts out of \(8\) are white, and those \(5\) parts are \(25\) beads.

\(1 \text{ part}\)	\(=\)	\(25 \div 5 = 5\)
\(\text{total}\)	\(=\)	\(8 \times 5 = \boldsymbol{40}\)
\(\text{blue}\)	\(=\)	\(3 \times 5 = \boldsymbol{15}\)

total = 40

Check: \(\dfrac{15}{40} = \dfrac{3}{8}\) ✓

Example 4 — Changing the probability

A jar holds \(5\) orange and \(15\) purple lollies. How many orange lollies must be added so that the probability of choosing an orange lolly is \(\dfrac{2}{5}\)?

Solution

Let \(k\) orange lollies be added. The orange count and the total both go up by \(k\); the purple count stays at \(15\).

\(\dfrac{5 + k}{20 + k}\)	\(=\)	\(\dfrac{2}{5}\)
\(5(5 + k)\)	\(=\)	\(2(20 + k)\)
\(25 + 5k\)	\(=\)	\(40 + 2k\)
\(3k\)	\(=\)	\(15\)
\(k\)	\(=\)	\(\boldsymbol{5}\)

k = 5

Check: \(\dfrac{10}{25} = \dfrac{2}{5}\) ✓. Add \(5\) orange lollies.

Common pitfalls

Use the total, not the other group. With \(4\) red and \(6\) blue, \(P(\text{red}) = \dfrac{4}{10}\), not \(\dfrac{4}{6}\).

1 is not a prime number. A prime has exactly two factors. \(1\) has only one, so leave it out when you count primes.

Adding items changes the total too. Adding \(3\) red to \(4\) red and \(6\) blue gives \(\dfrac{7}{13}\), not \(\dfrac{7}{10}\).

Simplify, then convert carefully. \(\dfrac{9}{12} = \dfrac{3}{4} = 0.75 = 75\%\). Round a percentage only at the end.

Expected is not certain. If you expect \(100\) results greater than \(4\) in \(300\) rolls, the actual count will usually be close to \(100\), but not exactly \(100\) every time.

Frequently asked questions

How do you calculate a probability?

When the outcomes are equally likely, divide the number of favourable outcomes by the total number of outcomes, then simplify.

What is a sample space?

It is the list of every possible outcome of a chance experiment. For one coin toss, \(S = \{\text{H}, \text{T}\}\).

What does 'at random' mean?

It means every item has the same chance of being chosen, so the outcomes are equally likely.

How do you write a probability as a percentage?

Multiply the fraction or decimal by \(100\). For example \(\dfrac{3}{8} = 0.375 = 37.5\%\).

How do you find the number of items from a probability?

Use the probability as parts of the whole. If \(5\) out of \(8\) parts are white and there are \(25\) white items, one part is \(5\) items and the whole is \(40\).

What is the expected number of times?

Multiply the probability by the number of trials. A fair coin tossed \(50\) times is expected to land on heads \(\dfrac{1}{2} \times 50 = 25\) times.

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Unions and Intersections

Review of Probability

📖 Theory

How to find a probability

How to work backwards

Complements and expected numbers

Common pitfalls

Frequently asked questions

Theory