Favourable out of possible

The idea

When all possible outcomes are equally likely, P(Event) = number of favourable outcomes ÷ number of possible outcomes; no trials or data are needed.

Reason it out

A die has 6 faces and one of them is a 4. If the die is fair, all six faces are equally likely. Can we find the chance of a 4 without rolling at all?

What fraction of the faces show a 4?

• 1 out of 6
• 1 out of 4
• 4 out of 6

Count favourable

Tap the favourable outcomes, the ones that make the event happen. Then Check and read the fraction.

Theory needs no dice

When all outcomes are equally likely: P(event) = number of favourable outcomes ÷ number of possible outcomes. No trials and no data are needed.

P(rolling a 4) = 1/6 ≈ 0.167. P(letter B in PROBABILITY) = 2/11 ≈ 0.182, because there are 2 Bs among 11 letters.

Notes

For equally likely outcomes: P(event) = favourable outcomes ÷ possible outcomes, with no experiment needed.

Check yourself

A letter is picked at random from PROBABILITY. What is P(B)?

How many outcomes are favourable for an even number when a die is rolled?

Answer: 3

The even numbers are 2, 4 and 6, so 3 outcomes. P(even) is 3/6, which is 1/2.

A die is rolled. What is P(a number less than 3), as a decimal rounded to two decimal places?

Answer: 0.33

Favourable: 1 and 2, so 2 of 6 outcomes. The fraction 2/6 is 0.333… ≈ 0.33.

Which of these do we not need to find a theoretical probability?

• 1/11. There are two Bs, so the favourable count is 2.
• 2/11 — correct. Yes. There are 2 Bs among 11 letters.
• 2/6. The word has 11 letters, not 6.
• The number of favourable outcomes. We need this.
• The number of possible outcomes. We need this too.
• A set of trials and their results — correct. Right. Theoretical probability only needs to count favourable and possible outcomes.