# 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 is equal to

## Solution :

Let the probability of getting a head be p and not getting a head be q.

Since, head appears first time in an even throw 2 or 4 or 6.

$$\therefore$$   $$2\over 5$$ = qp + $$q^3$$p + $$q^5$$p + ……

$$\implies$$  $$2\over 5$$ = $$qp\over {1 – q^2}$$

Since q = 1- p

$$\implies$$  $$2\over 5$$ = $$(1 – p)p\over {1 – (1 – p)^2}$$

$$\implies$$  $$2\over 5$$ = $$(1 – p)\over {2 – p}$$

$$\implies$$  4 – 2p = 5 – 5p

$$\implies$$  p =$$1\over 3$$

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