This one has left me a bit frustrated. I think there are $^{12}C_1 + ^{12}C_2 + ^{12}C_3 + ... + ^{12}C_5$ ways to get a $5$ card hand with at least one face card $(1585)$.
There should be $^{28}C_5$ ways to get a five card hand whose contents are all $\geq 8 (98280)$.
I need to divide the cardinality of the intersection by that of the sample space.
I was using a sample space of $^{52}C_5$, but should the sample space be smaller?
I'm getting a bit tangled up here.
On
Lets just use the normal Bayes formula:
$P(All \; cards \geq8|At \; least \; one \; face)= \frac{P(All cards \geq8,At \; least \; one \; face \;)}{P(At \; least \; one \; face \;)}$
You can quite easily compute $P(At \; least \; one \; face \;)$ by just taking the complement. The actual verification i leave to you
edit: As suggested, there was a serious typo( the events arent independent). The computation of the thing in the bracket is not evident, so what we can do is use the full Bayes formula: $P(All \; cards \geq8|At \; least \; one \; face)=\frac{P(At \; least \; one \; face|All \; cards \geq8)*P(All \; cards \geq8)}{P(At \; least \; one \; face)}$
This time around(hopefully), every term is easy to compute
By exchangeability of the cards (ie symmetry), without loss of generality we may assume that the first card is a face card ($J$, $Q$ or $K$). But then we have no information on the further ones so, so we just need them all to be at least an $8$, where we are drawing uniformly at random without replacement from a set of $51$ cards (not $52$, since we have removed a $J$/$Q$/$K$), where $6 \cdot 4 = 24$ of them are 'bad' (namely the $2$s, $3$s, $4$s, $5$s, $6$s and $7$s in the four suits -- I'm assuming $A$ce is high).
Thus the probability is simply $$ \left( 1 - \frac{24}{51} \right) \cdot \left( 1 - \frac{24}{50} \right) \cdot \left( 1 - \frac{24}{49} \right) \cdot \left( 1 - \frac{24}{48} \right). $$ Simplify/manipulate this expression as you desire.