A student is selected at random, and he is found to suffer from - CBSE Class 12 Sample Paper for 2026 Boards

part 2 - Question 38 (ii) - CBSE Class 12 Sample Paper for 2026 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 3 - Question 38 (ii) - CBSE Class 12 Sample Paper for 2026 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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Question 38 (ii) A student is selected at random, and he is found to suffer from anxiety and low retention issues. What is the probability that he/she spends screen time more than 4 hours per day?We need to find P(E1|A) Using conditional probability formula P(E1|A) = (š‘·(š‘¬_šŸāˆ© š‘Ø))/(š‘·(š‘Ø)) Now, we found P(A) in last part For š‘·(š‘¬_šŸāˆ© š‘Ø) can also write P(A|E1) = (š‘ƒ(š“ ∩ šø_1))/(š‘ƒ(šø_1)) Putting values 80/100=(š‘ƒ(š“ ∩ šø_1))/(60/100) 80/100 Ɨ60/100=š‘ƒ(š“ ∩ šø_1) š‘·(š‘Ø ∩ š‘¬_šŸ )=šŸ–šŸŽ/šŸšŸŽšŸŽ Ć—šŸ”šŸŽ/šŸšŸŽšŸŽ š‘ƒ(š“ ∩ šø_1 )=8/10 Ɨ6/10 š‘·(š‘Ø ∩ š‘¬_šŸ )=šŸ’šŸ–/šŸšŸŽšŸŽ Putting values in (1) P(E1|A) = (š‘ƒ(šø_1∩ š“))/(š‘ƒ(š“)) P(E1|A) = (48/100)/(72/100) P(E1|A) = šŸ’šŸ–/šŸ•šŸ P(E1|A) = 6/9 P(E1|A) = šŸ/šŸ‘ Thus, required probability is šŸ/šŸ‘

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 15 years. He provides courses for Maths, Science and Computer Science at Teachoo