MIP2602 Assignment 4 (COMPLETE ANSWERS) 2025 - DUE 15 August 2025

MIP2602 Assignment 4
(COMPLETE ANSWERS)
2025 - DUE 15 August
2025
NO PLAGIARISM
[Pick the date]
[Type the company name]


,Exam (elaborations)
MIP2602 Assignment 4 Memo | Due 15
August 2025
 Course
 Mathematics for Intermediate Phase teachers IV (MIP2602)
 Institution
 University Of South Africa (Unisa)

MIP2602 Assignment 4 Memo | Due 15 August 2025. Step by Step
Calculations provided. SECTION A: PROBABILITY QUESTION 1 1.1 A bag
contains a total of 10 marbles: 3 red marbles, 4 blue marbles, 2 green
marbles, and 1 yellow marble. 1.1.1 If one marble is randomly drawn from
the bag, what is the theoretical probability of drawing: 1.1.1.1 A red marble
= ____________ 1.1.1.2 A blue marble = ____________ 1.1.1.3 A green marble =
____________ 1.1.1.4 A yellow marble = ____________ (4) 1.1.2 Conduct an
experiment where you draw one marble at a time from the bag and record
the colour, returning it each time. Repeat this 14 times. Use the table below
to track your results: (7) 1.1.3 Based on the results you found in question
1.1.2, calculate the experimental probability for each colour: 1.1.3.1 P(Red)
= ____________ 1.1.3.2 P(Blue) = ____________ 1.1.3.3 P(Green) = ____________
1.1.3.4 P(Yellow) = ____________ (4) 1.1.4 How do the theoretical and
experimental probabilities compare in your results? Mention one reason why
they might be different. (2) 1.2 One letter is selected at random from the
word STATISTICS. What is the probability that the letter is: 1.2.1 S (2) 1.2.2 E
(2) 1.2.3 T or C (3) 1.3 A jar contains 4 red, 5 yellow, and 6 blue marbles.
Two marbles are drawn one after the other with replacement. Determine the
probability that: 1.3.1 Neither marble is red. (2) 1.3.2 The first is blue and the
second is yellow. (2)

✅ QUESTION 1
1.1.1 Theoretical Probability

There are 10 marbles total:

 3 Red
 4 Blue
 2 Green
 1 Yellow

The theoretical probability is:

, P(Event)=Number of favourable outcomesTotal number of outcomesP(\text{Event}) = \frac{\
text{Number of favourable outcomes}}{\text{Total number of outcomes}}
P(Event)=Total number of outcomesNumber of favourable outcomes

1.1.1.1 Red:

P(Red)=310P(\text{Red}) = \frac{3}{10}P(Red)=103

1.1.1.2 Blue:

P(Blue)=410=25P(\text{Blue}) = \frac{4}{10} = \frac{2}{5}P(Blue)=104=52

1.1.1.3 Green:

P(Green)=210=15P(\text{Green}) = \frac{2}{10} = \frac{1}{5}P(Green)=102=51

1.1.1.4 Yellow:

P(Yellow)=110P(\text{Yellow}) = \frac{1}{10}P(Yellow)=101

✅ [4 marks]



1.1.2 Experimental Probability Table (7 marks)

Let’s assume you did 14 trials and recorded the following counts:

Colour Frequency
Red 4
Blue 6
Green 3
Yellow 1
Total 14

(You can change this if your actual trial values differ.)



1.1.3 Experimental Probability

P(Colour)=FrequencyTotal number of trialsP(\text{Colour}) = \frac{\text{Frequency}}{\
text{Total number of trials}}P(Colour)=Total number of trialsFrequency

1.1.3.1 Red: