JEE Advance - Mathematics (2000 - No. 2)
A coin has probability $$p$$ of showing head when tossed. It is tossed $$n$$ times. Let $${p_n}$$ denote the probability that no two (or more) consecutive heads occur. Prove that $${p_1} = 1,{p_2} = 1 - {p^2}$$ and $${p_n} = \left( {1 - p} \right).\,\,{p_{n - 1}} + p\left( {1 - p} \right){p_{n - 2}}$$ for all $$n \ge 3.$$
The probability of no two consecutive heads for n=1 is 1.
The probability of no two consecutive heads for n=2 is 1-p^2.
The recurrence relation for pn is pn = (1-p)pn-1 + p(1-p)pn-2 for n>=3.
The base cases for the recurrence relation are p1 = 1 and p2 = 1 - p.
The recurrence relation accounts for the cases where the nth toss is tails or heads.
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