STA1501 ASSIGNMENT 2 (Unique Nr.: 888411) SEM 1 OF 2023 EXPECTED QUESTION AND ANSWERS.

Question 1

1) True
2) False
3) False
4) True
5) False
6) False
7) False
8) False
9) True
10) False



Question 1



(a)

(i) 𝑃(𝐴 or 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) - 𝑃(𝐴 and 𝐵) = (3/6) + (3/6) - (1/6) = 5/6

(ii) 𝑃(𝐴 or 𝐶) = 𝑃(𝐴) + 𝑃(𝐶) - 𝑃(𝐴 and 𝐶) = (3/6) + (2/6) - (0/6) = 5/6

(iii) 𝑃(𝐴 and 𝐵) = 𝑃(𝐴 ∩ 𝐵) = 𝑃({3}) = 1/6

(iv) 𝑃(𝐵 or 𝐶) = 𝑃(𝐵) + 𝑃(𝐶) - 𝑃(𝐵 and 𝐶) = (3/6) + (2/6) - (1/6) = 4/6

(v) 𝑃(𝐵|𝐶) = 𝑃(𝐵 and 𝐶)/𝑃(𝐶) = (1/6)/(2/6) = 1/2

(vi) 𝑃(𝐴|𝐵) = 𝑃(𝐴 and 𝐵)/𝑃(𝐵) = (1/6)/(3/6) = 1/3



(b) 𝐴 and 𝐵 are independent if 𝑃(𝐴 and 𝐵) = 𝑃(𝐴) × 𝑃(𝐵). We have 𝑃(𝐴 and 𝐵) = 1/6, 𝑃(𝐴) = 3/6, and
𝑃(𝐵) = 3/6. Thus, 𝑃(𝐴 and 𝐵) = (3/6) × (3/6) = 1/6, which is equal to 𝑃(𝐴) × 𝑃(𝐵). Therefore, 𝐴 and 𝐵 are
independent.



(c) 𝐵 and 𝐶 are mutually exclusive if 𝐵 ∩ 𝐶 = ∅. We have 𝐵 ∩ 𝐶 = {4, 5} ≠ ∅. Therefore, 𝐵 and 𝐶 are not
mutually exclusive.

, Question 2



(a)

(i) 𝑃(𝐶) = (Number of facilities in location C) / (Total number of facilities) = ≈ 0.4265

(ii) 𝑃(𝐵) = (Number of facilities in B) / (Total number of facilities) = ≈ 0.7574

(iii) 𝑃(𝐴 and 𝐶) = (Number of facilities in A and C) / (Total number of facilities) = ≈ 0.1176

(iv) 𝑃(𝐶 | 𝐴) = 𝑃(𝐴 and 𝐶) / 𝑃(𝐴) = () / () = ≈ 0.1203

(v) 𝑃(𝐵 or 𝐷) = 𝑃(𝐵) + 𝑃(𝐷) - 𝑃(𝐵 and 𝐷) = () + () - () ≈ 0.9257



(b) Two events are independent if the occurrence of one does not affect the probability of the
occurrence of the other. We can show that 𝐶 and 𝐷 are independent by verifying if 𝑃(𝐶 and 𝐷) = 𝑃(𝐶) ×
𝑃(𝐷).

𝑃(𝐶 and 𝐷) =

𝑃(𝐶) =

𝑃(𝐷) =

𝑃(𝐶 and 𝐷) = () × () = ≈ 0.1194



Since 𝑃(𝐶 and 𝐷) = 𝑃(𝐶) × 𝑃(𝐷), we can conclude that 𝐶 and 𝐷 are independent.



(c) Two events are mutually exclusive if they cannot occur together. We can show that 𝐴 and 𝐶 are
mutually exclusive by verifying if 𝑃(𝐴 and 𝐶) = 0.

𝑃(𝐴 and 𝐶) = ≈ 0.1176

Since 𝑃(𝐴 and 𝐶) ≠ 0, we can conclude that 𝐴 and 𝐶 are not mutually exclusive.