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A biased coin with probability p, 0<p<1 of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is 2/5, then p equals A. 1/3 B. 2/3 C. 2/5 D. 3/5 Select the correct answer from above options

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Correct Answer - A Let q=1-p. Since head appears first time in an even throw i.e., 2 or 4 or 6 etc. ∴ 2 5 =ap+q3p+q5p+...... ⇒ 2 5 = qp 1-q2 ⇒ 2 5 = (1-p)p 1-(1-p)2 = 1-p 2-p ⇒2(2-p)=5(1-p)⇒p=1/3

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