Sec 2 probability: count one event

For equally likely outcomes, count the outcomes that fit the event and divide by all possible outcomes. Keep repeated labels as separate equally likely sectors.

What this guide covers

Selected Sec 2 S2 probability of single events. Two-stage draws, tree diagrams and independent-event calculations are later work.

Count outcomes, not distinct labels

Eight equal sectors are eight possible outcomes even if several carry the same number. A probability must lie from 0 to 1; the event and its complement together cover all outcomes.

Repeated labels on a spinner

A fair spinner has eight equal sectors labelled 1,1,2,2,2,3,3,4. Find the probability of an even number and of a number greater than 2.

  1. There are eight equally likely sectors. Even labels occupy three 2-sectors and one 4-sector: 4 outcomes, so P(even)=48=12.
  2. Labels greater than 2 occupy two 3-sectors and one 4-sector: 3 outcomes, so P(>2)=38.

Check: The labels are not four equally likely numbers; the repeated sectors must each be counted.

Try it yourself

Work on paper, then compare the steps. These checks do not establish mastery.

A fair six-sided die is rolled once. What is the probability of a prime number?

Prime outcomes are 2, 3 and 5: 3 of 6 equally likely faces. Probability =36=12. The number 1 is not prime.

A box contains 5 green, 3 yellow and 2 blue identical counters. One is selected at random. Find the probability that it is not yellow.

Total =5+3+2=10. Not yellow means green or blue: 5+2=7 outcomes, so probability =710. Equivalently, 1−310=710.

Where this connects

Sources and scope

Original Mentora spinner and counter tasks aligned to selected Sec 2 S2 content in the MOE G2/G3 Mathematics syllabus. No third-party spinner graphic or question was used.

MOE G2 and G3 Mathematics syllabuses ↗