###### ICT

October 12, 2023###### Database-Management-System

October 12, 2023# Database-Management-System

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