Class 12 Probability · Short Answer2 Marksconditional_probability
Class 12 Maths · Probability · Short Answer #50
Given P(E)=0.8, P(F)=0.7 and P(E∩F)=0.6, find P(E′|F′).
✅ Step-by-Step Solution
Final Answer: 1/3
- P(E′∩F′)=1−P(E∪F).
- The union has probability 0.8+0.7−0.6=0.9, so the intersection of complements has probability 0.1. Also P(F′)=0.3.
- Thus P(E′|F′)=0.1/0.3=1/3.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] P(E′∩F′)=1−P(E∪F).
- [1 mark] The union has probability 0.8+0.7−0.6=0.9, so the intersection of complements has probability 0.1. Also P(F′)=0.3.
- [0.5 marks] Thus P(E′|F′)=0.1/0.3=1/3.
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