What this quiz covers

Grade 12 Counting & Probability Quiz 1 draws its questions at random from 34 questions across these topics:


Worked examples

A few of the question types, with full solutions, so you know what to expect.

The rules. For two events A and B, \(P(A)=0{,}4\), \(P(B)=0{,}5\) and \(P(A \text{ and } B)=0{,}2\). Determine \(P(A \text{ or } B)\).

  1. \(0{,}9\)
  2. \(0{,}7\)
  3. \(0{,}2\)
  4. \(1{,}1\)

Answer: \(0{,}7\)

Use the general addition rule, which is true for any two events:

\(P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\)

\(= 0{,}4 + 0{,}5 - 0{,}2\)

Tip: The overlap is subtracted once, otherwise it is counted in both P(A) and P(B).

The rules. Events A and B are mutually exclusive, with \(P(A)=0{,}25\) and \(P(B)=0{,}35\). Determine \(P(A \text{ or } B)\).

  1. \(0{,}60\)
  2. \(0{,}0875\)
  3. \(0{,}51\)
  4. \(0{,}10\)

Answer: \(0{,}60\)

Mutually exclusive means the events cannot both happen, so \(P(A \text{ and } B) = 0\).

\(P(A \text{ or } B) = 0{,}25 + 0{,}35 - 0\)

Tip: Only drop the subtraction when the question actually says mutually exclusive.

The rules. If \(P(A) = 0{,}72\), write down \(P(\text{not } A)\).

  1. \(0{,}72\)
  2. \(1{,}72\)
  3. \(0{,}28\)
  4. \(0{,}36\)

Answer: \(0{,}28\)

A and 'not A' are complementary: together they fill the whole sample space and cannot overlap.

\(P(\text{not } A) = 1 - P(A) = 1 - 0{,}72\)

The rules. A and B are independent events with \(P(A)=0{,}3\) and \(P(B)=0{,}6\). Determine \(P(A \text{ and } B)\).

  1. \(0{,}9\)
  2. \(0{,}18\)
  3. \(0\)
  4. \(0{,}3\)

Answer: \(0{,}18\)

For independent events the multiplication rule applies:

\(P(A \text{ and } B) = P(A) \times P(B) = 0{,}3 \times 0{,}6\)

Tip: Independent means MULTIPLY. Mutually exclusive means ADD. They are different ideas.

The rules. Given \(P(A)=0{,}6\), \(P(B)=0{,}3\) and \(P(A \text{ or } B)=0{,}72\). Are A and B independent?

  1. Yes, because \(P(A \text{ and } B) = P(A)\times P(B) = 0{,}18\)
  2. No, because \(P(A \text{ and } B) \neq 0\)
  3. Yes, because \(P(A) + P(B) > P(A \text{ or } B)\)
  4. It cannot be determined without more information

Answer: Yes, because \(P(A \text{ and } B) = P(A)\times P(B) = 0{,}18\)

First find the intersection using the addition rule:

\(0{,}72 = 0{,}6 + 0{,}3 - P(A \text{ and } B)\)

\(\therefore P(A \text{ and } B) = 0{,}9 - 0{,}72 = 0{,}18\)

Now test independence: \(P(A)\times P(B) = 0{,}6 \times 0{,}3 = 0{,}18\)

Tip: To prove independence you must show both sides separately, then state that they are equal.