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

Last updated at September 2, 2025 by Teachoo

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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Transcript

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) = (𝑷(𝑬_𝟏∩ 𝑨))/(𝑷(𝑨)) Now, we found P(A) in last part For 𝑷(𝑬_𝟏∩ 𝑨) can also write P(A|E1) = (𝑃(𝐴 ∩ 𝐸_1))/(𝑃(𝐸_1)) Putting values 80/100=(𝑃(𝐴 ∩ 𝐸_1))/(60/100) 80/100 ×60/100=𝑃(𝐴 ∩ 𝐸_1) 𝑷(𝑨 ∩ 𝑬_𝟏 )=𝟖𝟎/𝟏𝟎𝟎 ×𝟔𝟎/𝟏𝟎𝟎 𝑃(𝐴 ∩ 𝐸_1 )=8/10 ×6/10 𝑷(𝑨 ∩ 𝑬_𝟏 )=𝟒𝟖/𝟏𝟎𝟎 Putting values in (1) P(E1|A) = (𝑃(𝐸_1∩ 𝐴))/(𝑃(𝐴)) P(E1|A) = (48/100)/(72/100) P(E1|A) = 𝟒𝟖/𝟕𝟐 P(E1|A) = 6/9 P(E1|A) = 𝟐/𝟑 Thus, required probability is 𝟐/𝟑

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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

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