ICT
October 12, 2023DatabaseManagementSystem
October 12, 2023DatabaseManagementSystem
Question 221

820

Explanation :
Probability of 1st condition being satisfied(say P(A)) = 10/15 = 2/3
Probability of 2nd condition being satisfied(say P(B)) = 1/20
Probability of both conditions being satisfied(say P(A intersection B)) = 2/3*1/20 = 1/30
Probability of any one condition being satisfied = P(A union B) = P(A)+P(B)P(A intersection B) = 2/3 + 1/20 – 1/30 = 41/60
therefore, expected number of tuples = (41/60)*1200 = 820
Explanation :
Probability of 1st condition being satisfied(say P(A)) = 10/15 = 2/3
Probability of 2nd condition being satisfied(say P(B)) = 1/20
Probability of both conditions being satisfied(say P(A intersection B)) = 2/3*1/20 = 1/30
Probability of any one condition being satisfied = P(A union B) = P(A)+P(B)P(A intersection B) = 2/3 + 1/20 – 1/30 = 41/60
therefore, expected number of tuples = (41/60)*1200 = 820