InterviewSolution
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InterviewSolution
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In a game a coin is tossed `2n+m`times and a player wins if he does not get any two consecutive outcomessame for at least `2n`times in a row. The probability that player wins the game is`(m+2)/(2^(2n)+1)`b. `(2n+2)/(2^(2n))`c. `(2n+2)/(2^(2n+1))`d. `(m+2)/(2^(2n))`A. `(m+2)/(2^(2n)+1)`B. `(2n+2)/(2^(2n))`C. `(2n+2)/(2^(2n+1))`D. `(m+2)/(2^(2n))` |
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Answer» Correct Answer - D Player should get `(HT, HT, HT,...)orTH, TH,...)` at least 2n times. If the sequence starts from first place, then the probability is `1//2^(2n)` and if starts from any other place, then the probability is `1//2^(2n+1).` Hence, required probability is `2((1)/(2^(2n))+(m)/(2^(2n+1)))=(m+32)/(2^(2n))` |
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