GO Classes Test Series 2024 | Mock GATE | Test 12 | Question: 6

Answer 1

The probability of no conflict is $\dfrac{10 \cdot 9 \cdot 8}{10^3}=0.72$. So the probability of there being at least one scheduling conflict is $0.28.$

Comments

can we not do this from binomial ?

10C3 (0.3)^3(0.7)^7 = 0.27

@Sachin Mittal 1 @GO Classes @Deepak Poonia

Answer 2

Let Set of courses = C = {C1, C2, C3}
Let Time Interval = T = {T1, T2, T3,……., T9, T10}
Let’s solve this problem by complement logic i.e. probability of no two course should map to same time interval, which is nothing but probability of one-one function from Set C to Set T.
no of one-one function = 10P3 = 720
total function = 10^3 = 1000
P(one -one ) = 720/1000 = 0.72;
hence, probability of two course map to same time interval = 1 – 0.72 = 0.28