Die A has 4 red and 2 white faces, whereas die B has 2 red and 4 white faces. A coins 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 32/33 when it is given that red turns up every time in first `n` throws, then find the value of `ndot`

Explanation

Let R be the event 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
`thereforeP((R)/(E_(1)))=((2)/(3))^(n)and P((R)/(E_(2)))=((1)/(3))^(n)`
As per the given condition,
`(P(E_(1))P(R//E_(1)))/(P(E_(1))P(R//E_(1))+P(E_(2))P(R//E_(2)))=32/33`
`or (1/2((2)/(3))^(n))/(1/2((2)/(3))^(v)+1/2((1)/(3))^(n))=32/33`
or `(2^(n))/(2^(n)+1)=32/33`
or n = 5