Masters of International Business: PM, 1st year
Distance learning: Quantitative Methods
Darja Žiļcova

Practical exercises Topic 1

1. If positive answer is YES, but negative answer is NO
S = { (YES,NO), (YES,YES), (NO,YES), (NO, NO) }
So there are 4 sample points
2. A) Formula for rolling a pair of dice sample points: S=(n6) (n6) = 6x6=36
B) List of sample points:
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
C) Probability of obtaining a value of 7 is
6/36 = 1/6
D) Probability of obtaining a value of 9 and greater is
10/36 = 5/18
E) No, not agree! There are 18/36 = 1/2 probabilities of even values and the same
18/36 = 1/2 probabilities of odd values
F) Counting rule for multi step eksperiments
3. If 5 outcomes are equally likely, it means that 1/5 = 0.2
A) P(A) = E1 + E2 = 0.2 + 0.2= 0.4
P(B) = E3+ E4 = 0.2 + 0.2= 0.4
P(C) = E2+ E3 + E5 = 0.2 + 0.2 + 0.2 = 0.6
B) P(AUB) = .4+.4 = .8, mutually exclusive
C) Ā = {E3, E4, E5} notC = {E1,E4}
P(Ā) = 1 – 0.4 = 0.6

P(notC) = 1 – 0.6 = 0.4
D) notB = {E1, E2, E5}
A U notB = {E1, E2} + {E1, E2, E5}
P(notB) = 1 – 0.4 = 0.6
P(A U notB) = P(A) + P(notB) = 0.4 + 0.6 = 1
E) P(B U C) = P(B) + P(C) = 0.4 + 0.6 = 1
4. P(personal U business) = P(personal) + P(business) - P(personal and business) =
= 0.54 + 0.458 – 0.3 = 0.698 = 69.8%
Complement: P = 1 - P(personal U business) = 0.302 = 30.2%
5. A) P(A and B) = 0, because A and B are mutually exclusive
B) P(A|B) = 0, because A happens, B can’t happen – they are mutually exclusive
C) Not agree .
In this example A and B are mutually exclusive, but they are still dependent. Since we are sure A can’t happen if we know B happens and vice versa. If they were independent, we would have P(A|B) = P(A) = 0.3
D) A and B are depended, because one can’t happen at the same time with another
– this is rule of mutually exclusive events.
6.1 A) Was used formula P (A|B) = P(A and B) / P(B)
In our case P(B|A1) = P(B and A1) / P(A1), so to find P(B and A1) we have to put known values and calculate P(B and A1) = 0.5 * 0.2 = 0.1
And the same with anothers:
P(B and A2) = 0.4 * 0.5 = 0.2
P(B and A3) = 0.3 * 0.3 = 0.9
B) Bayes theorem:
P(A2|B) = P(A2)*P(B|A2) / [P(A1)*P(B|A1) + P(A2)*P(B|A2) + P(A3)*P(B|A3)] =
= 0.5*0.2/(0.2*0.1 + 0.5*0.2 + 0.3*0.9) = 0.175
P(A1|B) = P(A1)*P(B|A1) / [P(A1)*P(B|A1) + P(A2)*P(B|A2) + P(A3)*P(B|A3)] =
= 0.2*0.1/(0.2*0.1 + 0.5*0.2 + 0.3*0.9) = 0.351
P(A3|B) = P(A3)*P(B|A3) / [P(A1)*P(B|A1) + P(A2)*P(B|A2) + P(A3)*P(B|A3)] =
= 0.3*0.9/(0.2*0.1 + 0.5*0.2 + 0.3*0.9) = 0.474
C) P(A1|B) = P(A1)*(P(B|A1) = 0.2*0.1 = 0.2
P(A2|B) = P(A2)*(P(B|A2) = 0.5*0.2=0.1
P(A3|B) = P(A3)*(P(B|A3) = 0.3*0.3=0.9

6.2 D1 – customers default
D2 – Customers don’t default
M – Missed Payments
P(D1) = 0.05
P(D2) = 0.95
P(M|D2) = 0.2 (missing a monthly payment)
P(M|D1) = 1 (who default)
A) P(D1|M) = 0.5*1 / 0.5*1 + 0.95*0.2 = 0.5/0.24=0.21
B) 0.21>0.2 so the probability of default is greater than 0.2