When we perform an experiment (like tossing a coin or rolling a die), all possible outcomes together form the sample space, denoted by $S$.

An event is a specific outcome or set of outcomes we are interested in. It is a subset of the sample space, denoted by $E$.

Example 1

A fair die is rolled once.

• Sample space: $S = \{1, 2, 3, 4, 5, 6\}$
• Event $E$ = rolling an even number: $E = \{2, 4, 6\}$
• The complement of $E$ (not getting an even number): $E' = \{1, 3, 5\}$

The probability of an event $E$ is calculated using the formula:

$P(E) = \frac{n(E)}{n(S)}$

Where:

• $n(E)$ = number of favorable outcomes
• $n(S)$ = total number of possible outcomes in the sample space

Key property: $0 \leq P(E) \leq 1$

Also, $P(E') = 1 - P(E)$ where $E'$ is the complement of $E$ (the event that $E$ does not occur).

Worked Example 2

A box contains 5 red pens and 3 blue pens. One pen is drawn at random. Find the probability that the pen is red.

Solution:

• Total pens = $5 + 3 = 8$
• Number of red pens = 5

$P(\text{Red}) = \frac{5}{8}$

Worked Example 3

What is the probability of NOT getting a 4 when a fair die is rolled?

Solution:

• Sample space: $S = \{1, 2, 3, 4, 5, 6\}$
• Probability of getting a 4: $P(4) = \frac{1}{6}$
• Probability of NOT getting a 4: $P(\text{not } 4) = 1 - \frac{1}{6} = \frac{5}{6}$

Two events are mutually exclusive (or disjoint) if they cannot occur at the same time. For example, when rolling a die, you cannot get both an even number and the number 3 at the same time.

If $A$ and $B$ are mutually exclusive events:

$P(A \text{ or } B) = P(A \cup B) = P(A) + P(B)$

Worked Example 4

A card is drawn from a standard deck. Find the probability of drawing a king or a queen.

Solution:

• Number of kings = 4
• Number of queens = 4
• These events are mutually exclusive (a card cannot be both king and queen)

$P(\text{king or queen}) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$

Two events are independent if the occurrence of one does not affect the occurrence of the other. For independent events $A$ and $B$:

$P(A \text{ and } B) = P(A \cap B) = P(A) \times P(B)$

Dependent events occur when the outcome of the first event affects the probability of the second event (usually without replacement).

Worked Example 5 (Independent Events)

A coin is tossed and a die is rolled. Find the probability of getting a tail and a number greater than 4.

Solution:

• $P(\text{tail}) = \frac{1}{2}$
• $P(\text{number} > 4) = \frac{2}{6} = \frac{1}{3}$ (numbers 5 and 6)

$P(\text{tail and number} > 4) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$

Worked Example 6 (Dependent Events)

A bag contains 6 green apples and 4 red apples. Two apples are drawn one after another without replacement. Find the probability that both are green.

Solution:

• First apple green: $P(G_1) = \frac{6}{10} = \frac{3}{5}$
• Second apple green (after removing one green apple): $P(G_2) = \frac{5}{9}$

$P(\text{both green}) = \frac{3}{5} \times \frac{5}{9} = \frac{3}{9} = \frac{1}{3}$

A tree diagram helps visualize all possible outcomes of combined events. Each branch shows a step in the experiment, and we multiply probabilities along the branches.

Worked Example 7

A family has two children. Draw a tree diagram and find the probability that both children are girls. (Let G = girl, B = boy)

Solution:

Tree diagram:

• First child: G ($\frac{1}{2}$) or B ($\frac{1}{2}$)
• Second child: for each first-child outcome, G ($\frac{1}{2}$) or B ($\frac{1}{2}$)

Sample space: $\{(G,G), (G,B), (B,G), (B,B)\}$

$P(\text{both girls}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Worked Example 8 (Tanzanian Context)

A small shop in Arusha sells mandazi (mandazi) and mkate (bread). The probability that they have mandazi in stock is $\frac{3}{4}$, and the probability they have mkate is $\frac{2}{3}$ (these are independent). Find the probability that they have both items in stock.

Solution:

$P(\text{both}) = \frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}$

Two-way tables help organize information when two events are considered together.

Worked Example 9

In a class of 30 students, 18 like Mathematics (M), 15 like Physics (P), and 8 like both. If a student is selected at random, find the probability that the student likes Mathematics or Physics.

Solution:

Using the formula for union of two events:

$P(M \cup P) = P(M) + P(P) - P(M \cap P)$

$P(M \cup P) = \frac{18}{30} + \frac{15}{30} - \frac{8}{30} = \frac{25}{30} = \frac{5}{6}$

Probability is used daily in Tanzania, for example, when shopping at a local market in Dar es Salaam. If a vendor has 20 mangoes and 5 are overripe, the probability of randomly picking a good mango is $\frac{15}{20} = \frac{3}{4}$. This helps buyers make informed choices. Similarly, weather forecasts use probability to predict whether it will rain, allowing farmers to plan their planting activities accordingly.