Question

A biased coin with probability of heads $p(0<p<1)$, 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 $3p$ is equal to

Answer 1

Let the probability of head is $p$.
The coin is tossed untill first head appears.
Let $P\left(E\right)=P\left(\text{number of tosses required is even}\right)=\frac{2}{5}$
Let there be $r$ tosses of coin before we get our first head.
$P\left(r\right)=(1-p{)}^{r-1}p$
$P\left(E\right)=P(r=2)+P(r=4)+\cdots$
$\Rightarrow P\left(E\right)=(1-p)p+(1-p{)}^{3}p+(1-p{)}^{5}p+\cdots$
$\Rightarrow P\left(E\right)=\frac{(1-p)p}{1-(1-p{)}^2}=\frac{2}{5}\Rightarrow 5p-5p^2=4p-2p^2\therefore 5-5p=4-2p(\because p>0)\Rightarrow 3p=1$