Question

I alternately toss a fair coin and throw a fair die, until I, either toss a head or throw the face two. If I toss the coin first, then the probability that I throw the face two before I toss a head, is

• A. $\frac{1}{7}$
• B. $\frac{7}{12}$
• C. $\frac{5}{12}$
• D. $\frac{5}{7}$

Answer 1

The correct option is A $\frac{1}{7}$

Let $H$ be tossing a head.
$P\left(H\right)=\frac{1}{2}$
and $A$ be tossing a $2$ with die.
$P\left(A\right)=\frac{1}{6}$
$E:$ Tossing the face two before tossing a head
$P\left(E\right)=P\left(\overline{H}\right)\cdot P\left(A\right)+P\left(\overline{H}\right)\cdot P\left(\overline{A}\right)\cdot P\left(\overline{H}\right)\cdot P\left(A\right)+P\left(\overline{H}\right)\cdot P\left(\overline{A}\right)\cdot P\left(\overline{H}\right)\cdot P\left(\overline{A}\right)\cdot P\left(\overline{H}\right)\cdot P\left(A\right)+\cdots$
$\Rightarrow P\left(E\right)=\frac{1}{2}\cdot \frac{1}{6}+\frac{1}{2}\cdot \frac{5}{6}\cdot \frac{1}{2}\cdot \frac{1}{6}+\cdots$
$\Rightarrow P\left(E\right)=\frac{1}{12}+\frac{5}{12}\cdot \frac{1}{12}+{\left(\frac{5}{12}\right)}^2\cdot \frac{1}{12}+\cdots$
$\Rightarrow P\left(E\right)=\frac{\frac{1}{12}}{1-\frac{5}{12}}=\frac{1}{7}$