Theoretical probability. A bag holds 3 red, 4 blue and 7 black pens. Find \(P(\text{red})\).

1. \(\dfrac{3}{14}\)
2. \(\dfrac{3}{11}\)
3. \(\dfrac{1}{3}\)
4. \(\dfrac{3}{7}\)

Answer: \(\dfrac{3}{14}\)

Count the pens: \(3 + 4 + 7 = 14\) altogether.

Three of them are red.

Tip: The denominator is the TOTAL, not the other colours.

Theoretical probability. A bag holds 3 red, 4 blue and 7 black pens. Find \(P(\text{black})\).

1. \(\dfrac{1}{2}\)
2. \(\dfrac{7}{10}\)
3. \(\dfrac{1}{7}\)
4. \(\dfrac{2}{7}\)

Answer: \(\dfrac{1}{2}\)

Seven of the 14 pens are black.

\(\dfrac{7}{14} = \dfrac{1}{2}\)

Theoretical probability. A die is rolled once. Find \(P(\text{a natural number})\).

1. \(1\)
2. \(0\)
3. \(\dfrac{1}{6}\)
4. \(\dfrac{5}{6}\)

Answer: \(1\)

Every face (1 to 6) is a natural number, so all 6 outcomes are favourable.

\(\dfrac{6}{6} = 1\)

Tip: A probability of 1 means the event is CERTAIN.

Theoretical probability. A die is rolled once. Find \(P(\text{a number greater than } 6)\).

1. \(0\)
2. \(1\)
3. \(\dfrac{1}{6}\)
4. \(\dfrac{1}{3}\)

Answer: \(0\)

No face on a six-sided die is bigger than 6, so 0 outcomes are favourable.

\(\dfrac{0}{6} = 0\)

Tip: A probability of 0 means the event is IMPOSSIBLE.

Theoretical probability. Which value can NOT be a probability?

1. \(\dfrac{7}{6}\)
2. \(0\)
3. \(0{,}45\)
4. \(1\)

Answer: \(\dfrac{7}{6}\)

Every probability satisfies \(0 \leq P \leq 1\).

\(\dfrac{7}{6}\) is greater than 1.