Question

A biased coin (with probability of obtaining a head equal to $p>0$) is tossed repeatedly and independently until the first head is observed. The probability that the first head appears at an even numbered toss, is

• A. $\frac{1-p}{2-p}$
• B. $\frac{2-p}{1-p}$
• C. $\frac{p}{2-p}$
• D. $\frac{p}{1-p}$

Answer 1

The correct option is A $\frac{1-p}{2-p}$

Let,
Event $H_1:$ head on first toss
Event $E:$ first head on even numbered toss
By total probability theorem:
$P\left(E\right)=P\left(E|H_1\right)P\left(H_1\right)+P\left(E|H_1^{\prime }\right)P\left(H_1^{\prime }\right)$
Now clearly, $P\left(E|H_1\right)=0$
$P\left(E|H_1^{\prime }\right)=P\left(E^{\prime }\right)=1-P\left(E\right)$
$\therefore P\left(E\right)=0\times p+(1-P(E\left)\right)\times (1-p)$
$\Rightarrow P\left(E\right)=\frac{1-p}{2-p}$