MIP2602 Assignment 4 (750826) Due 15 August 2025

MIP2602
ASSIGNMENT 04
UNIQUE NO: 750826
DUE 15 AUGUST 2025

,MIP2602

ASSIGNMENT 04

DUE 15 AUGUST 2025

UNIQUE NUMBER: 750826




SECTION A: PROBABILITY



QUESTION 1

1.1.1 Theoretical Probability Calculation

Let the total number of marbles in the bag be:

𝑇𝑜𝑡𝑎𝑙 = 3 (𝑅𝑒𝑑) + 4(𝐵𝑙𝑢𝑒) + 2(𝐺𝑟𝑒𝑒𝑛) + 1(𝑌𝑒𝑙𝑙𝑜𝑤) = 10Total
= 3(Red) + 4(Blue) + 2(Green) + 1(Yellow) = 10𝑇𝑜𝑡𝑎𝑙
= 3(𝑅𝑒𝑑) + 4(𝐵𝑙𝑢𝑒) + 2(𝐺𝑟𝑒𝑒𝑛) + 1(𝑌𝑒𝑙𝑙𝑜𝑤) = 10

The theoretical probability P(E)P(E)P(E) of an event is given by:

P(E)=Number of favorable outcomes
Number of favorable outcomes
Total number of outcomes P(E) = Total number of outcomes
𝑃(𝐸) =

𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠

Thus:

• 1.1.1.1 P(Red marble)

3
𝑃(𝑅𝑒𝑑) = 310𝑃(Red) = 𝑃(𝑅𝑒𝑑) = 103
10

• 𝟏. 𝟏. 𝟏. 𝟐𝑷(𝑩𝒍𝒖𝒆𝒎𝒂𝒓𝒃𝒍𝒆)

4
𝑃(𝐵𝑙𝑢𝑒) = 410𝑃(Blue) = 𝑃(𝐵𝑙𝑢𝑒) = 104
10

, • 𝟏. 𝟏. 𝟏. 𝟑𝑷(𝑮𝒓𝒆𝒆𝒏𝒎𝒂𝒓𝒃𝒍𝒆)

2
𝑃(𝐺𝑟𝑒𝑒𝑛) = 210𝑃(Green) = 𝑃(𝐺𝑟𝑒𝑒𝑛) = 102
10

• 𝟏. 𝟏. 𝟏. 𝟒𝑷(𝒀𝒆𝒍𝒍𝒐𝒘𝒎𝒂𝒓𝒃𝒍𝒆)

1
𝑃(𝑌𝑒𝑙𝑙𝑜𝑤) = 110𝑃(Yellow) = 𝑃(𝑌𝑒𝑙𝑙𝑜𝑤) = 101
10



1.1.3 Experimental Probability

The experimental probability is calculated using the formula:

P(E)=Number of times event occurredTotal number of trialsP(E) = \frac{\text{Number of
times event occurred}}{\text{Total number of
trials}}P(E)=Total number of trialsNumber of times event occurred

Assume the results from 1.1.2 (as the user has not provided the tally):

Colour Drawn Frequency (Assumed Example)

Red 5

Blue 4

Green 3

Yellow 2

Total 14



• 𝟏. 𝟏. 𝟑. 𝟏𝑷(𝑹𝒆𝒅)

5
𝑃(𝑅𝑒𝑑) = 514𝑃(Red) = 𝑃(𝑅𝑒𝑑) = 145
14

• 𝟏. 𝟏. 𝟑. 𝟐𝑷(𝑩𝒍𝒖𝒆)

4
𝑃(𝐵𝑙𝑢𝑒) = 414𝑃(Blue) = 𝑃(𝐵𝑙𝑢𝑒) = 144
14