Probability Problems
Probability Problems in NSW Year 8 maths (NESA outcome MA4‑PRO‑C‑01) takes the basic rule as known and focuses on careful counting: large or unusual sample spaces, events described with “and”, “or” and “not”, and objects that repeat, such as a die with faces 1, 1, 2, 3, 3, 3.
You will count runs of numbers, multiples and square numbers, work out chances with a full deck of cards or the days of a month, recount after some items are removed, and see that a fair coin does not remember earlier tosses. You will also work backwards from a decimal probability to a count, and compare games at a show or tickets in four raffles to find the best chance. It is part of Year 8 Probability.
Every question with a fully worked solution.
- Probability Problems - Video - Probability of Simple Event Watch
Theory
Most probability problems are counting problems: find the total number of equally likely outcomes, count the ones the words describe, then divide. From \(1\) to \(21\) there are \(8\) primes, so \(P(\text{not prime}) = \dfrac{13}{21}\). This Year 8 maths page shows how to count carefully.
The rule. When all outcomes are equally likely, \(P(\text{event}) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}\), and \(P(\text{not A}) = 1 - P(\text{A})\).
Counting a run of numbers. From \(12\) to \(40\) there are \(40 - 12 + 1 = 29\) numbers: subtract, then add \(1\) for the first one. The multiples of \(7\) up to \(50\): \(50 \div 7\) is \(7\) remainder \(1\), so there are \(7\).
“Or” lists. Count each list, then subtract the outcomes that are in both, so nothing is counted twice. “And” means an outcome must pass both tests.
Count objects, not labels. A die with faces \(2, 2, 2, 5, 5, 8\) has \(6\) equally likely faces, so \(P(2) = \dfrac{3}{6} = \dfrac{1}{2}\), not \(\dfrac{1}{3}\). The same goes for coins in a jar and sectors on a spinner.
When the sample changes, recount both numbers before dividing: the favourable count and the total.
Every trial is new. A fair coin, die or spinner does not remember earlier results, so the probability is the same on every trial.
Working backwards. number \(=\) \(P \times\) total, and total \(=\) number \(\div P\). With \(P(\text{win}) = 0.2\) and \(9\) winning tickets there are \(9 \div 0.2 = 45\) tickets.
Comparing chances. Write each probability over a common denominator, or as a decimal or percentage, then compare.
Read the words, then decide what to count:
| The words | What to count |
|---|---|
| \(A\) and \(B\) | outcomes that pass both tests |
| \(A\) or \(B\) | both lists, shared outcomes counted once |
| not \(A\) | every outcome except those in \(A\) |
| at least \(8\) | \(8\) and above |
| at most \(8\) | \(8\) and below |
| greater than \(8\) | \(9\) and above |
| less than \(8\) | \(7\) and below |
| A standard deck of 52 cards | Count |
|---|---|
| 4 suits: hearts, diamonds (red); clubs, spades (black) | \(13\) cards in each suit |
| red cards; black cards | \(26\) each |
| each rank (ace, 2, …, 10, jack, queen, king) | \(4\) of each |
| picture cards (jack, queen, king) | \(12\) |
Solving a probability problem
- Find the total number of equally likely outcomes (objects, faces, sectors or numbers).
- Read the words (and, or, not, at least, greater than) and write the first and last values.
- List or count the favourable outcomes; for “or”, subtract the overlap.
- Divide and simplify, or convert to a decimal or percentage if asked.
- Recount both numbers if anything is removed or added.
Working backwards from a probability
- Number \(=\) probability \(\times\) total.
- Total \(=\) number \(\div\) probability.
- The rest \(=\) total \(-\) number, and \(P(\text{rest}) = 1 - P\).
(i) red
(ii) not white
(iii) green or white.
| \(\text{Total}\) | \(=\) | \(3 + 2 + 4 = 9\) |
| \(P(\text{red})\) | \(=\) | \(\dfrac{4}{9}\) |
| \(P(\text{not white})\) | \(=\) | \(1 - \dfrac{2}{9} = \dfrac{7}{9}\) |
| \(P(\text{green or white})\) | \(=\) | \(\dfrac{3 + 2}{9} = \dfrac{5}{9}\) |
(i) \(\dfrac{4}{9}\)
(ii) \(\dfrac{7}{9}\)
(iii) \(\dfrac{5}{9}\)
(i) not prime
(ii) even
(iii) a multiple of \(3\).
Primes: \(2, 3, 5, 7, 11, 13, 17, 19, 23\) (nine). Even: every second number from \(2\) up to \(24\) (twelve). Multiples of \(3\): \(3, 6, 9, 12, 15, 18, 21, 24\) (eight).
| \(P(\text{not prime})\) | \(=\) | \(\dfrac{25 - 9}{25} = \dfrac{16}{25}\) |
| \(P(\text{even})\) | \(=\) | \(\dfrac{12}{25}\) |
| \(P(3\text{s})\) | \(=\) | \(\dfrac{8}{25}\) |
(i) \(\dfrac{16}{25}\)
(ii) \(\dfrac{12}{25}\)
(iii) \(\dfrac{8}{25}\)
(i) \(P(\text{a multiple of } 9)\)
(ii) \(P(\text{a multiple of } 6 \text{ or } 15)\)
(iii) \(P(\text{not a multiple of } 10)\).
(i) Multiples of \(9\): \(9, 18, 27, 36, 45, 54\), six of them.
(ii) \(60 \div 6 = 10\) multiples of \(6\) and \(60 \div 15 = 4\) multiples of \(15\); \(30\) and \(60\) are in both.
(iii) There are six multiples of \(10\).
| \(P(9\text{s})\) | \(=\) | \(\dfrac{6}{60} = \dfrac{1}{10}\) |
| \(\text{Count}\) | \(=\) | \(10 + 4 - 2 = 12\) |
| \(P(6 \text{ or } 15)\) | \(=\) | \(\dfrac{12}{60} = \dfrac{1}{5}\) |
| \(P(\text{not } 10\text{s})\) | \(=\) | \(\dfrac{60 - 6}{60} = \dfrac{9}{10}\) |
(i) \(\dfrac{1}{10}\)
(ii) \(\dfrac{1}{5}\)
(iii) \(\dfrac{9}{10}\)
(i) \(P(7)\)
(ii) \(P(\text{odd})\)
(iii) \(P(9)\) on the next roll, after it has just shown \(9\) three times in a row.
The six faces are equally likely, so the total is \(6\).
| \(P(7)\) | \(=\) | \(\dfrac{2}{6} = \dfrac{1}{3}\) |
| \(P(\text{odd})\) | \(=\) | \(\dfrac{3}{6} = \dfrac{1}{2}\) |
| \(P(9)\) | \(=\) | \(\dfrac{1}{6}\) |
(i) \(\dfrac{1}{3}\)
(ii) \(\dfrac{1}{2}\) (the two 7s and the 9)
(iii) \(\dfrac{1}{6}\): the next roll is a new trial, so three 9s in a row change nothing.
(i) Find \(P(\text{berry})\).
(ii) Two apricot and three choc bars are eaten. Find \(P(\text{apricot})\) now.
(iii) Find \(P(\text{choc})\) now, as a percentage.
| \(P(\text{berry})\) | \(=\) | \(\dfrac{7}{30}\) |
| \(\text{New total}\) | \(=\) | \(30 - 5 = 25\) |
| \(P(\text{apricot})\) | \(=\) | \(\dfrac{12 - 2}{25} = \dfrac{2}{5}\) |
| \(P(\text{choc})\) | \(=\) | \(\dfrac{11 - 3}{25} = \dfrac{32}{100}\) |
(i) \(\dfrac{7}{30}\)
(ii) \(\dfrac{2}{5}\)
(iii) \(32\%\). Both the count and the total changed.
(i) How many counters are in the bag?
(ii) How many are purple?
(iii) Green counters are added until \(P(\text{green}) = 0.4\). How many were added?
| \(\text{Total}\) | \(=\) | \(21 \div 0.35 = 60\) |
| \(\text{Purple}\) | \(=\) | \(60 - 21 = 39\) |
| \(\text{New total}\) | \(=\) | \(39 \div 0.6 = 65\) |
| \(\text{Green added}\) | \(=\) | \(65 - 39 - 21 = 5\) |
(i) \(60\)
(ii) \(39\)
(iii) \(5\). Purple stays at \(39\), and \(P(\text{purple}) = 1 - 0.4 = 0.6\). Check: \(26 \div 65 = 0.4\).
(ii) A whole number from \(1\) to \(50\) is chosen at random. Find, as a decimal, the probability that it contains the digit \(3\).
(i) The six numbers: \(257, 275, 527, 572, 725, 752\). Four start with \(5\) or \(7\).
(ii) Ones digit \(3\): \(3, 13, 23, 33, 43\) (five). Tens digit \(3\): \(30\) to \(39\) (ten). \(33\) is in both lists.
| \(P(\text{over } 500)\) | \(=\) | \(\dfrac{4}{6} = \dfrac{2}{3}\) |
| \(\text{Count}\) | \(=\) | \(5 + 10 - 1 = 14\) |
| \(P(\text{a } 3)\) | \(=\) | \(\dfrac{14}{50} = 0.28\) |
(i) \(\dfrac{2}{3}\)
(ii) \(0.28\)
| \(P(A)\) | \(=\) | \(\dfrac{5}{120} \approx 4.2\%\) |
| \(P(B)\) | \(=\) | \(\dfrac{3}{60} = 5\%\) |
| \(P(C)\) | \(=\) | \(\dfrac{9}{200} = 4.5\%\) |
Raffle B gives the best chance, even though Mia holds the fewest tickets there.
Common pitfalls
Frequently asked questions
How do I count the whole numbers from 12 to 40?
Subtract and add 1: 40 − 12 + 1 = 29 numbers.
How do I find the probability of A or B?
Count the outcomes in A and in B, subtract those in both so they are counted once, then divide by the total.
Does a coin that has landed heads four times become more likely to land tails?
No. A fair coin has no memory, so the probability of a tail is still 1/2.
What is the difference between at least 8 and greater than 8?
At least 8 includes 8; greater than 8 starts at 9.
How do I find the number of items from a probability?
Multiply the probability by the total. To find the total, divide the number of items by the probability.
How do I decide which game gives the better chance?
Write each probability as a decimal or a percentage, or over a common denominator, and choose the largest.