Given three probabilities $Pr(\beta|\alpha)$, $Pr(\gamma|\alpha)$ and $Pr(\alpha)$; and that $\beta$ and $\gamma$ are conditionally independent given $\alpha$, can $Pr(\alpha|\beta, \gamma)$ be calculated?
My solution:
$$\begin{align} Pr(\alpha|\beta, \gamma) &= \frac{Pr(\alpha, \beta, \gamma)}{Pr(\beta, \gamma)} \\ &= \frac{Pr(\alpha) Pr(\beta, \gamma|\alpha)}{Pr(\beta, \gamma)} \end{align}$$
Since $\beta$ and $\gamma$ are independent given $\alpha$, we can get:
$Pr(\alpha|\beta, \gamma) = \frac{Pr(\alpha) Pr(\beta|\alpha) Pr(\gamma|\alpha)}{Pr(\beta, \gamma)}$
Three probabilities in the numerator are given but the denominator is not, then the given constraints are not sufficient to calculate $Pr(\alpha|\beta, \gamma)$.
Am I right?
You could use Bayes Theorem and then use the independence given $\alpha$: $$\mathbb{P}(\beta, \gamma) = \sum_{\alpha} \mathbb{P}(\beta, \gamma | \alpha)\mathbb{P}(\alpha)= \sum_{\alpha}\mathbb{P}(\beta | \alpha) \mathbb{P} (\gamma | \alpha) \mathbb{P} (\alpha) $$
P.S.: I'm assuming you are given those probabilities for each $\alpha$ and $\alpha$ spans in a discrete set.