Question

A biased coin with probability p, \(0<\mathrm{p}<1\), of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is \(\frac{2}{5}\), then p is equal to :

• A. \(\frac{1}{2}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{2}{3}\)

Hint:

The number of tosses is even if the first head comes on toss 2, 4, 6, ...; sum the resulting infinite geometric series with \(q=1-p\).

Answer 1

The Correct Option is B

Let \(q=1-p\). The first head appears on an even toss 2, 4, 6, ... with probabilities \(qp,\ q^3p,\ q^5p,\dots\).
P(even) \(=qp\,(1+q^2+q^4+\dots)=\frac{qp}{1-q^2}=\frac{qp}{(1-q)(1+q)}=\frac{q}{1+q}\) (since \(1-q=p\)).
Given \(\frac{q}{1+q}=\frac{2}{5}\Rightarrow 5q=2+2q\Rightarrow q=\frac{2}{3}\).
Hence \(p=1-q=\frac{1}{3}\).
Answer: (2) \(\frac{1}{3}\)