Functional Skills Level 2 — Probability

Calculating Probability

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The Basic Probability Formula

When all outcomes are equally likely, probability is simply the number of favourable outcomes divided by the total number of possible outcomes.

\[P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of equally-likely outcomes}}\]
Standard Dice — P(4) = 1 ÷ 6 Favourable favourable 1 6 total outcomes = ¹⁄₆ ≈ 0.167 Only one face shows 4 — so P(4) = ¹⁄₆ Bag of Counters — Calculating Each Probability R R R B B B B G G G Total counters = 10 P(red) = 3/10 = 0.3 P(blue) = 4/10 = 0.4 P(green) = 3/10 = 0.3 Check: 0.3+0.4+0.3 = 1 ✓

Example 1 — Probability from a dice

Step 1
A fair dice has 6 faces: 1, 2, 3, 4, 5, 6. All equally likely.
Step 2
Find \(P(\text{even number})\). Even numbers: 2, 4, 6 — that is 3 favourable outcomes.
Step 3
\(P(\text{even}) = \frac{3}{6} = \frac{1}{2} = 0.5\)

Example 2 — Probability from a bag of counters

Step 1
A bag has 5 red, 3 blue, 2 yellow counters. Total = 10.
Step 2
Find \(P(\text{blue})\). Favourable = 3.
Step 3
\(P(\text{blue}) = \frac{3}{10} = 0.3\)
Step 4
Check: \(P(\text{red}) + P(\text{blue}) + P(\text{yellow}) = \frac{5}{10} + \frac{3}{10} + \frac{2}{10} = 1\) ✓
All probabilities for an experiment must add up to 1. Use this as a check after listing all outcomes.
✓ Quick Check 1 — The Basic Probability Formula
Question 1 of 10
Complement and Multiple Outcomes

You met the complement rule (\(P(\text{event}) + P(\text{not event}) = 1\)) in Lesson 14.1 — here are two more applications of it, plus a new rule for combining outcomes. The complement of an event is everything that is NOT that event: sometimes it's easier to calculate \(P(\text{not A})\) and subtract from 1. We can also add probabilities when there are multiple favourable outcomes.

Spinner — Multiple Favourable Outcomes 8 equal sections: 3 red → P(red) = 3/8 3 blue → P(blue) = 3/8 2 green → P(green) = 2/8 P(red or blue) = 3/8 + 3/8 = 6/8 = 3/4

Example 3 — Using the complement

Step 1
A bag has 10 counters: 7 red, 3 blue. Find \(P(\text{not red})\).
Step 2
Method 1 (direct): \(P(\text{not red}) = \frac{3}{10} = 0.3\)
Step 3
Method 2 (complement): \(P(\text{not red}) = 1 - \frac{7}{10} = \frac{3}{10} = 0.3\) ✓

Example 4 — Quality control context

Step 1
A factory makes light bulbs. Inspection shows that \(P(\text{faulty}) = 0.03\).
Step 2
\(P(\text{not faulty}) = 1 - 0.03 = 0.97\)
Step 3
If 5000 bulbs are made, expected faulty = \(0.03 \times 5000 = \mathbf{150}\)
Multiple outcomes: When the event includes several outcomes (e.g. "red or blue"), add their individual probabilities — but only if the outcomes are mutually exclusive (they can't both happen at the same time).
✓ Quick Check 2 — Complement and Multiple Outcomes
Question 1 of 10
AND / OR Rules — Independent and Mutually Exclusive Events

When two events are involved, you need two key rules. AND (both happen — multiply). OR (at least one happens — add, if they can't both happen at the same time).

\[P(A \text{ AND } B) = P(A) \times P(B) \quad \text{(when A and B are independent)}\] \[P(A \text{ OR } B) = P(A) + P(B) \quad \text{(when A and B are mutually exclusive)}\]
Venn Diagram — Mutually Exclusive vs Overlapping Events A B Mutually exclusive: No overlap — cannot both happen A B Both Overlapping: can both happen AND Rule — Two Independent Events (Two Coins) H P(H) = 1/2 × H P(H) = 1/2 = P(HH) = ½ × ½ = ¼ = 0.25 Multiply probabilities for AND (independent events)

Example 5 — AND rule (independent events)

Step 1
A coin is flipped and a dice is rolled. What is \(P(\text{heads AND 6})\)?
Step 2
The coin and dice are independent — the result of one doesn't affect the other.
Step 3
\(P(\text{heads}) = \frac{1}{2}, \quad P(6) = \frac{1}{6}\)
Step 4
\(P(\text{heads AND 6}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \approx 0.083\)

Example 6 — OR rule (mutually exclusive events)

Step 1
A dice is rolled. What is \(P(\text{1 OR 6})\)?
Step 2
Rolling a 1 and rolling a 6 cannot happen at the same time — mutually exclusive.
Step 3
\(P(1) = \frac{1}{6}, \quad P(6) = \frac{1}{6}\)
Step 4
\(P(\text{1 OR 6}) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \approx 0.333\)
Summary: AND = multiply (independent events). OR = add (mutually exclusive events). These rules appear frequently in FS Level 2 probability questions.
✓ Quick Check 3 — AND / OR Rules
Question 1 of 10

Interactive — Spinner Simulator

Number of sectors: 4
Sector Expected probability Times landed Actual %

Practice: Calculating Probability

Type your answer as a decimal (e.g. 0.25) or a fraction (e.g. 1/4), then click Check.

#QuestionAnswerResult
1A fair dice is rolled. Find \(P(\text{rolling a 5})\).
2A bag has 4 red and 6 blue counters. Find \(P(\text{red})\).
3A spinner has 10 equal sections: 6 green, 4 orange. Find \(P(\text{orange})\).
4\(P(\text{rain tomorrow}) = 0.35\). Find \(P(\text{not rain})\).
5A factory finds \(P(\text{faulty part}) = 0.02\). Find \(P(\text{not faulty})\).
6A bag has 3 red, 2 blue, 5 green counters (10 total). Find \(P(\text{red or blue})\).
7A bag has 12 numbered tickets, 1 to 12. Find \(P(\text{picking a multiple of 3})\).
8A dice is rolled. Find \(P(1 \text{ or } 6)\).
9A spinner has 8 equal sections numbered 1 to 8. Find \(P(\text{a number less than 3})\).
10In a survey, \(P(\text{person prefers tea}) = 0.65\). Find \(P(\text{person does not prefer tea})\).

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