In a school, for a period of 8 working days, it is equally likely that Tanishka is present or…

For each of the 8 days, there are 2 outcomes (Present P or Absent A). Total possible outcomes = $2^8 = 256$. We find the number of sequences with at least 5 consecutive present days (P): Case 1: Exactly 5 consecutive P's starting on day 1: PPPPP A _ _ (the 6th day must be A to avoid double counting, unless all are P). Let us list all disjoint patterns that contain at least 5 consecutive P's: - Starting at day 1: PPPPP _ _ _ gives $2^3 = 8$ sequences. - Starting at day 2: APPPPP _ _ gives $2^2 = 4$ sequences. - Starting at day 3: _ APPPPP _ gives $2^2 = 4$ sequences. - Starting at day 4: _ _ APPPPP gives $2^2 = 4$ sequences. Total favourable outcomes = $8 + 4 + 4 + 4 = 20$. Therefore, the probability is $\frac{20}{256} = \frac{5}{64}$.