- Messages
- 330
- Reaction score
- 191
- Points
- 53
Re:
P (either A or B) = P(A) + P(B) - P(A intersection B)
= 0.3 + 0.6 - 0.18 = 0.72
P (both A and B) = P( A intersection B ) = 0.18
P (either A or B but not both occurs) = 0.72 - 0.18 = 0.54
aleezay said:nov07P2Q2(ii)
P (either A or B) = P(A) + P(B) - P(A intersection B)
= 0.3 + 0.6 - 0.18 = 0.72
P (both A and B) = P( A intersection B ) = 0.18
P (either A or B but not both occurs) = 0.72 - 0.18 = 0.54