Probability – dice, tree diagrams and examples

When outcomes are equally likely

If all possible outcomes are equally likely, count the favourable outcomes and divide by the number of possible outcomes.

A fair six-sided die has outcomes 1, 2, 3, 4, 5 and 6. The event «an even number» has three favourable outcomes: 2, 4 and 6. The probability is therefore 3/6 = 1/2 = 50%.

Probabilities lie between zero and one. Zero means impossible in the model and one means certain. In more general situations, counting outcomes is not enough; you need to know their probabilities.

Want to switch between fractions and percentages? Revise calculating with fractions and percentages.

Two independent events

For independent events, one event does not affect the probability of the other. You find the probability that both occur by multiplying.

You toss a fair coin and roll a fair die, independently. The probability of heads is 1/2 and of an even number on the die is 1/2. The probability of both is 1/2 · 1/2 = 1/4.

An earlier coin toss does not change the probability of the next independent toss. Getting tails several times does not make heads more likely next time.

Tree diagrams: with and without replacement

A bag contains three red and two blue balls. You draw one ball at random and then another. Without replacement, four balls remain after the first draw.

• First ball red: 3/5.
  • Then red: 2/4. Both red: 3/5 · 2/4 = 3/10.
  • Then blue: 2/4. Red then blue: 3/10.
• First ball blue: 2/5.
  • Then red: 3/4. Blue then red: 3/10.
  • Then blue: 1/4. Both blue: 1/10.

The list shows the branches of a tree diagram. Multiply the probabilities along a branch. Add branch probabilities when several branches match what you are looking for.

With replacement, put the first ball back and mix before the next draw. The probability of red is then 3/5 again, and two red balls gives 3/5 · 3/5 = 9/25.

At least one: use the complementary event

Sometimes it is easier to calculate what must not happen. The complement of «at least one head in two coin tosses» is «no heads», meaning two tails.

For two independent tosses of a fair coin, the probability of two tails is 1/2 · 1/2 = 1/4. Therefore, the probability of at least one head is 1 − 1/4 = 3/4.

Practise probability with answers

Problem 1

What is the probability of rolling a number greater than four on a fair six-sided die?

Problem 2

Two balls are drawn without replacement from the bag above. What is the probability of one ball of each colour?

Problem 3

What is the probability of at least one six in two independent rolls of a fair six-sided die?