CAIE A-Level Mathematics 5.3 Probability Question Bank

Practise building probability models from tables, trees and counting, then applying addition, multiplication, complements, conditional probability and independence tests.

Exam points

translate the outcome structure into a tree, table or counting model before assigning probabilities
add mutually exclusive routes and multiply successive probabilities, adjusting for replacement
use conditional fractions or compare P(A ∩ B) with P(A)P(B) to test independence

Covered syllabus points

5.3.1—Permutations and combinations 5.3.2—Probability rules 5.3.3—Exclusive and independent events 5.3.4—Elementary probability

[Maximum number: 4]

Raman and Sanjay are members of a quiz team which has 9 members in total. Two photographs of the quiz team are to be taken.

For the first photograph, the 9 members will stand in a line.

For a random division into a group of 5 and a group of 4, find the probability that Raman and Sanjay are in the same group as each other.

[Number of ways with Raman and Sanjay together on back row = ] $^{7} \mathrm{C}_{3}$
[Number of ways with Raman and Sanjay together on front row = ] $^{7} \mathrm{C}_{2}$
M1
${ }^{7} \mathrm{C}_{x}$ seen, x=3 or 2.
[Total =] 35+21
M1
Summing two correct scenarios.
56
A1
Evaluated - may be seen used in probability.
If M0 scored, SC B1 for 56 WWW.
Probability $=\frac{\text { their } 56}{\text { their }(c)}=\frac{56}{126}, \frac{4}{9}, 0.444$
B1 FT
FT their 56 from adding 2 or more scenarios in numerator and their (c) or correct as denominator.

[Maximum number: 2]

The residents of Mahjing were asked to classify their local bus service:
- 25 % of residents classified their service as good.
- 60 % of residents classified their service as satisfactory.
- 15 % of residents classified their service as poor.

Three residents of Mahjing are selected at random.

Find the probability that one resident classified the bus service as good, one as satisfactory and one as poor.

The three classifications can occur in 3!=6 orders. Hence,
$P=0.25(0.6)(0.15)\times6=0.135.$

[Maximum number: 10]

Bag A contains 4 balls numbered 2, 4, 5, 8. Bag B contains 5 balls numbered 1, 3, 6, 8, 8. Bag C contains 7 balls numbered 2,7,8,8,8,8,9. One ball is selected at random from each bag.
- Event X is 'exactly two of the selected balls have the same number'.
- Event Y is 'the ball selected from bag A has number 4'.

(a)

Find P(X).

[ 5 ]

\mathrm{P}(X)=\mathrm{P}( exactly 2 balls have same number )
\mathrm{P}(2, N 2,2)=\frac{1}{4} \times 1 \times \frac{1}{7}=\frac{1}{28}
M1
Considering at least two options of 2s and 8s
$\mathrm{P}(8,8, \mathrm{~N} 8)=\frac{1}{4} \times \frac{2}{5} \times \frac{3}{7}=\frac{3}{70}$
M1
Considering three options for the 8s
$\mathrm{P}(8, \mathrm{~N} 8,8)=\frac{1}{4} \times \frac{3}{5} \times \frac{4}{7}=\frac{3}{35}$
M1
Summing their options if more than 3 in total
$\mathrm{P}(\mathrm{N} 8,8,8)=\frac{3}{4} \times \frac{2}{5} \times \frac{4}{7}=\frac{6}{35}$
B1
One option correct
\mathrm{P}(X)= sum $=\frac{47}{140}(0.336)$
A1

(b)

Find $\mathrm{P}(X \cap Y)$ and hence determine whether or not events X and Y are independent.

[ 3 ]

$\mathrm{P}(X \cap \mathrm{Y})=\mathrm{P}(4,8,8)=\frac{1}{4} \times \frac{2}{5} \times \frac{4}{7}=\frac{2}{35}$
B1
\mathrm{P}(Y)=\frac{1}{4}
\frac{2}{35} \neq \frac{47}{140} \times \frac{1}{4}
M1
Attempt to compare $\mathrm{P}(X \cap \mathrm{Y})$ with $\mathrm{P}(X) \times \mathrm{P}(Y)$ or
using conditional probabilities
Not independent
A1
Correct answer, correct working only

(c)

Find the probability that two balls are numbered 2, given that exactly two of the selected balls have the same number.

[ 2 ]

$P(\text{two balls numbered 2}\mid X) =\frac{1/28}{47/140}=\frac5{47}\approx0.106.$

All question bank results loaded

Covered syllabus points

5.3 Probability question 1

5.3 Probability question 2

5.3 Probability question 3

Question (a)

Question (b)

Question (c)