Question

Die A has 4 red and 2 white faces, whereas die B has 2 red and 4 white faces. A coin is flipped once. If it shows a head, the game continues by throwing die A; if it shows tail, then die B is to be used. If the probability that die A is used is $\frac{32}{33}$ when it is given that red turns up every time in first n throws, then find the value of n.

• A. n=5
• B. n=3
• C. n=4
• D. n=1

Answer 1

The correct option is A n=5

Let R be the even that a red face appears in each of the first n throws.
$E_1$: Die A is used when head has already fallen
$E_2$: Die B is used when tail has already fallen
$\therefore P\left(\frac{R}{E_1}\right)={\left(\frac{2}{3}\right)}^{n}andP\left(\frac{R}{E_2}\right)={\left(\frac{1}{3}\right)}^n$
As per the given condition,
$\frac{P\left(E_1\right)P\left(\frac{R}{E_1}\right)}{P\left(E_1\right)P\left(\frac{R}{E_1}\right)+P\left(E_2\right)P\left(\frac{R}{E_2}\right)}=\frac{32}{33}or\frac{\frac{1}{2}({\frac{2}{3}}^n}{)}\frac{1}{2}{\left(\frac{2}{3}\right)}^n+\frac{1}{2}{\left(\frac{1}{3}\right)}^n=\frac{32}{33}or\frac{2^n}{2^n+1}=\frac{32}{33}orn=5$