I have a question which is for some reason is not making sense in the results I get -
90% of candidates to be a president are not fit. There are three interviews to become a president. In each interview there is 70% to be rejected if you're not a fit and 60% to be accepted if you're a fit. all the interviews are not depended and the final decision is made by the majority. What is probability that someone is a fit if it was decided to reject him?
So I know it's a probability with a condition: (to be fit| to be rejected). I did a tree of probabilities and tried to put it in the formula - P(to be fit and to be rejected) / P(to be rejected).
So my three was starting at to be fit (0.1) and not to fit (0.9). Each branch was to be accepted if you're a fit (0.6), to be rejected if you're a fit (0.4), to be rejected if you're not a fit (0.7) and to be accepted if you're not a fit (0.3).
Since it's not depended I multiplied the results of the three interviews. I understood that since it's decided by the majority, you need to be rejected twice or all three times.
The tree is getting messy, am I missing something and the solution is simplier?
Thanks for the help
The majority thing can be dealt with separately from the conditional probability calculation: The probability that an unfit candidate is rejected by a majority is $\binom32\cdot0.7^2\cdot0.3+\binom33\cdot0.7^3=0.784$, and the probability that a fit candidate is accepted by a majority is $\binom32\cdot0.6^2\cdot0.4+\binom33\cdot0.6^3=0.648$.
Now we can calculate the conditional probability unencumbered by voting procedures:
\begin{eqnarray} \mathsf P(\text{fit}\mid\text{rejected}) &=& \frac{\mathsf P(\text{rejected}\mid\text{fit})\mathsf P(\text{fit})}{\mathsf P(\text{rejected})} \\ &=& \frac{\mathsf P(\text{rejected}\mid\text{fit})\mathsf P(\text{fit})}{\mathsf P(\text{rejected}\mid\text{fit})\mathsf P(\text{fit})+\mathsf P(\text{rejected}\mid\text{unfit})\mathsf P(\text{unfit})} \\ &=& \frac{(1-0.648)(1-0.9)}{(1-0.648)(1-0.9)+0.784\cdot0.9} \\ &=& \frac{22}{463} \\ &\approx& 4.75\%\;. \end{eqnarray}