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A fair die is rolled repeatedly. The probability that the cumulative sum is at least 17 in the third trial is

Solution

Correct Option: 1

Cumulative sum after 33 rolls is S3S_3. Need P(S317)P(S_3 \geq 17).

Maximum possible is 1818 (all sixes), so S3{17,18}S_3 \in \{17, 18\}.


S3=18S_3 = 18: outcome must be (6,6,6)(6, 6, 6), giving 11 way.

S3=17S_3 = 17: one die shows 55, the other two show 66. Choose which die shows 55: 33 ways.


Total favourable outcomes =1+3=4= 1 + 3 = 4

Total outcomes in 33 rolls =63=216= 6^3 = 216


Required probability ==

4216=154\dfrac{4}{216} = \dfrac{1}{54}

Answer =154= \dfrac{1}{54}

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