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 August 14, 2026 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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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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