Probability of the n-th person to believe the rumour The probability that a randomly selected person will believe a rumour about the extramarital affair of a politician is $0.6$. People from the general public are randomly selected one after the other independently. It is known that only one person between the first two selected believes the rumour, what is the probability that the sixth person to hear the rumour will be the third to believe approximately.
I don't understand why the answer is $^3C_2\times(0.6^2)\times(0.4^2)$.
 A: Among the first two people, one believes the rumour and one does not. So if the sixth person believes the rumour and is the third to do so, then among the third to fifth persons, one believes the rumour and two do not.
The probability that exactly one believes the rumour among the third to fifth persons is $\binom32$ (for selecting which persons do not believe it) times $0.4×0.4×0.6$ (the individual probabilities of believing the rumour or not). Then multiply this with the $0.6$ probability of the sixth person believing the rumour for a final answer of $\binom32(0.6)^2(0.4)^2$.
A: Imagine the possible scenarios:
$B, N, *, *, *, B\space$ or $N, B, *, *, *, B$.
For the first two persons, it is certain that one will believe and the other will not.
Now, $* = \left\{
\begin{array}{ll}
      B & p=0.6\\
      N & p=0.4\\
\end{array} 
\right. $
Since, we need exactly one of the $*$ to hold the value $B$, we get $3\choose2$ possibilties. 
Hence, required probability is ${3\choose 2} \times (0.6) \times (0.4)^2 \times (0.6)$
