Unit 16 Review Exercises

Probability

These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed. AHL marks questions (or parts) for HL only.

1 Paper 1 Two fair six-sided dice are rolled. Find the probability that

(a) the sum is a prime number,

(b) the product is a multiple of \(3\).

(a) Prime sums \(2, 3, 5, 7, 11\) occur \(1 + 2 + 4 + 6 + 2 = 15\) times: \(\;\dfrac{15}{36} = \dfrac{5}{12}\).

(b) The product is not a multiple of 3 when neither die shows 3 or 6: \(4 \times 4 = 16\) outcomes. \(\;\mathrm{P} = 1 - \dfrac{16}{36} = \dfrac{5}{9}\). See 16.1.

2 Paper 1 A spinner is spun \(200\) times and lands on red \(46\) times.

(a) Estimate the probability that it lands on red.

(b) Find the expected number of reds in \(350\) further spins.

(a) \(\dfrac{46}{200} = 0.23\).

(b) \(350 \times 0.23 = 80.5\). See 16.1.

3 Paper 1 \(A\) and \(B\) are independent events with \(\mathrm{P}(A) = 0.3\) and \(\mathrm{P}(B) = 0.5\). Find

(a) \(\mathrm{P}(A \cup B)\), \(\qquad\) (b) \(\mathrm{P}(A \mid B')\).

(a) \(0.3 + 0.5 - 0.3 \times 0.5 = 0.65\).

(b) \(A\) and \(B'\) are also independent, so \(\mathrm{P}(A \mid B') = \mathrm{P}(A) = 0.3\). See 16.2.

4 Paper 1 \(\mathrm{P}(A) = 0.2\) and \(\mathrm{P}(B) = k\). Find \(k\) if \(\mathrm{P}(A \cup B) = 0.5\) and

(a) \(A\) and \(B\) are mutually exclusive,

(b) \(A\) and \(B\) are independent.

(a) \(0.2 + k = 0.5\), so \(k = 0.3\).

(b) \(0.2 + k - 0.2k = 0.5\), so \(0.8k = 0.3\) and \(k = 0.375\). See 16.2.

5 Paper 1 The table shows how 120 visitors to a museum bought their tickets.

Online At the door Total
Adult 45 35 80
Child 15 25 40
Total 60 60 120

A visitor is chosen at random. Find

(a) \(\mathrm{P}(\text{online} \mid \text{child})\),

(b) \(\mathrm{P}(\text{adult} \mid \text{at the door})\).

(c) Are “bought online” and “child” independent? Give a reason.

(a) \(\dfrac{15}{40} = \dfrac{3}{8}\).

(b) \(\dfrac{35}{60} = \dfrac{7}{12}\).

(c) \(\mathrm{P}(\text{online}) = \dfrac{60}{120} = \dfrac{1}{2} \neq \dfrac{3}{8} = \mathrm{P}(\text{online} \mid \text{child})\), so they are not independent. See 16.2.

6 Paper 1 \(\mathrm{P}(A) = 0.4\), \(\mathrm{P}(B) = 0.45\) and \(\mathrm{P}(A' \cap B') = 0.3\).

(a) Draw a Venn diagram showing the probability of each region.

(b) Find \(\mathrm{P}(A \mid B)\).

(a) \(\mathrm{P}(A \cup B) = 1 - 0.3 = 0.7\), so \(\mathrm{P}(A \cap B) = 0.4 + 0.45 - 0.7 = 0.15\). Regions: \(A\) only \(0.25\), both \(0.15\), \(B\) only \(0.3\), neither \(0.3\).

(b) \(\mathrm{P}(A \mid B) = \dfrac{0.15}{0.45} = \dfrac{1}{3}\). See 16.4 and 16.2.

7 Paper 1 A bag contains 4 red and 2 green balls. Two balls are taken at random without replacement. Find the probability that

(a) the balls are different colours,

(b) the second ball is green.

(a) \(\mathrm{P}(RG) + \mathrm{P}(GR) = \dfrac{4}{6} \cdot \dfrac{2}{5} + \dfrac{2}{6} \cdot \dfrac{4}{5} = \dfrac{16}{30} = \dfrac{8}{15}\).

(b) \(\mathrm{P}(RG) + \mathrm{P}(GG) = \dfrac{8}{30} + \dfrac{2}{6} \cdot \dfrac{1}{5} = \dfrac{10}{30} = \dfrac{1}{3}\). See 16.4.

8 Paper 1 The probability that Zara catches the early bus is \(p\). If she catches it, the probability that she arrives on time is \(0.95\); otherwise it is \(0.6\). The probability that she arrives on time is \(0.81\).

(a) Find \(p\).

(b) AHL Given that she arrives on time, find the probability that she caught the early bus.

(a) \(0.95p + 0.6(1 - p) = 0.81 \Rightarrow 0.35p = 0.21 \Rightarrow p = 0.6\). See 16.4.

(b) \(\dfrac{0.6 \times 0.95}{0.81} = \dfrac{0.57}{0.81} = \dfrac{19}{27}\). See 16.3.

9 Paper 2 AHL Three technicians repair phones. \(T_1\) does \(40\%\) of the repairs, \(T_2\) does \(35\%\) and \(T_3\) does \(25\%\). The percentages of their repairs that are returned as faulty are \(5\%\), \(3\%\) and \(8\%\) respectively. A repaired phone is returned as faulty. Find the probability that it was repaired by \(T_3\).

\(\mathrm{P}(R) = 0.4 \times 0.05 + 0.35 \times 0.03 + 0.25 \times 0.08 = 0.02 + 0.0105 + 0.02 = 0.0505\).

\(\mathrm{P}(T_3 \mid R) = \dfrac{0.02}{0.0505} \approx 0.396\). See 16.3.

10 Paper 1 AHL A group of 3 is chosen at random from 4 teachers and 5 students. Find the probability that

(a) the group contains at least one teacher,

(b) the group contains exactly one teacher, given that it contains at least one teacher.

Total \(\dbinom{9}{3} = 84\). No teachers: \(\dbinom{5}{3} = 10\).

(a) \(1 - \dfrac{10}{84} = \dfrac{74}{84} = \dfrac{37}{42}\).

(b) Exactly one teacher: \(\dbinom{4}{1}\dbinom{5}{2} = 40\). \(\;\mathrm{P} = \dfrac{40}{74} = \dfrac{20}{37}\). See 16.5.