Question

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 $p=$........

• A. $2/5$
• B. $2/3$
• C. $1/3$
• D. $3/5$

Answer 1

The correct option is C $1/3$

Let $x$ denote the number of tosses required.
Then $P(x=r)={(1-P)}^{r-1}P;r=1,2,\mathrm{3....}$
Let $E$ denote the event that the number of tosses required is even. Then,
$P\left(E\right)=P\{(x=2)\cup (x=4)\cup (x=6)\cup ...\}=P(x=2)+P(x=4)+P(x=6)+....=p(1-p)+p{(1-p)}^3+p{(1-p)}^5+..$
$=\frac{p(1-p)}{1-{(1-p)}^2}=\frac{p(1-p)}{2p-p^2}=\frac{1-p}{2-p}$
As we are given that $P\left(E\right)=\frac{2}{5}$, we get
$5(1-p)=2(2-p)\Rightarrow 1-3p=0\Rightarrow p=\frac{1}{3}$