Q19·Section A (Section A)·19 of 60Probability of E without F or GLet E,FE,FE,F and GGG be any three events with P(E)=0.3P(E)=0.3P(E)=0.3, P(F∣E)=0.2P(F|E)=0.2P(F∣E)=0.2, P(G∣E)=0.1P(G|E)=0.1P(G∣E)=0.1 and P(F∩G∣E)=0.05P(F\cap G|E)=0.05P(F∩G∣E)=0.05. Then P(E−(F∪G))P(E-(F\cup G))P(E−(F∪G)) equals OptionsA0.1550.1550.155B0.1750.1750.175C0.2250.2250.225D0.2550.2550.255Pick an option to check your answer.Show answerShow solution Report a problem with this question← Q18All questionsQ20 →Keyboard: A–D answer · S solution · ←/→ previous/next