Calculating Basic Probability

Grade 3

The key formula

When all outcomes are equally likely, probability is simply a count:

P(event) = number of favourable outcomes ÷ total number of outcomes

Favourable outcomes just means the outcomes you're interested in — the ones where the event happens. Count those, divide by the total, and you're done.

A worked example

EXAMPLE

A standard dice is rolled once. Find: (a) P(4), (b) P(even number), (c) P(greater than 4).

A standard dice has 6 equally likely outcomes: 1, 2, 3, 4, 5, 6. (a) Only one face shows 4: P(4) = 1/6. (b) Even numbers are 2, 4, 6 — that's 3 outcomes: P(even) = 3/6 = 1/2. (c) Greater than 4 means 5 or 6 — that's 2 outcomes: P(>4) = 2/6 = 1/3.

Probabilities must add to 1

If you've found the probability of an event happening, you can get the probability of it not happening with one step:

P(event does not happen) = 1 − P(event happens)

This is called the complement. It's a quick check too — if P(A) + P(not A) ≠ 1, you've gone wrong somewhere.

EXAMPLE

The probability that it rains on a given day is 0.35. What is the probability that it does not rain?

P(no rain) = 1 − 0.35 = 0.65.

When outcomes aren't equally likely

The formula P = favourable ÷ total only works when every outcome has the same chance. If a spinner has sections of different sizes, or a biased coin is involved, you can't just count outcomes — you need the probabilities given in the question.

Try this one

A factory tests \( 1500 \) light bulbs. \( 36 \) of them are found to be faulty.

(a) Work out the relative frequency of a bulb being faulty. Give your answer as a decimal. [2]

(b) The factory produces \( 20\,000 \) bulbs in a week. Estimate how many will be faulty. [2]

Nova's hint:

This uses relative frequency as an estimate of probability, then applies it to a larger batch.

Ask yourself: 'what fraction of the tested bulbs were faulty, and how do I scale that up?' Once you have the relative frequency, multiply it by the larger total.

Try to Solve it with Nova?