Class 12 Probability · Very Long Answer5 Marksindependence
Class 12 Maths · Probability · Very Long Answer #119
Independent A,B satisfy P(A′∩B)=2/15 and P(A∩B′)=1/6. Find all possible ordered pairs (P(A),P(B)).
✅ Step-by-Step Solution
Final Answer: (1/5,1/6) or (5/6,4/5)
- Write x=P(A), y=P(B). Independence gives (1−x)y=2/15 and x(1−y)=1/6.
- Subtract these equations: y−x=2/15−1/6=−1/30, so y=x−1/30.
- Substitute in x(1−y)=1/6: x(31/30−x)=1/6, giving 30x²−31x+5=0.
- Factor (5x−1)(6x−5)=0. Thus x=1/5 or5/6, with y=1/6 or4/5 respectively.
- Both pairs are valid probabilities and satisfy both given products: (1/5,1/6) and (5/6,4/5).
📋 CBSE Stepwise Marking Scheme
- [1 mark] Write x=P(A), y=P(B). Independence gives (1−x)y=2/15 and x(1−y)=1/6.
- [1 mark] Subtract these equations: y−x=2/15−1/6=−1/30, so y=x−1/30.
- [1 mark] Substitute in x(1−y)=1/6: x(31/30−x)=1/6, giving 30x²−31x+5=0.
- [1 mark] Factor (5x−1)(6x−5)=0. Thus x=1/5 or5/6, with y=1/6 or4/5 respectively.
- [1 mark] Both pairs are valid probabilities and satisfy both given products: (1/5,1/6) and (5/6,4/5).
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