What is the reason of the low probability of a B-meson decaying into two muons I found a Nature article where they measure the probability of a B-meson decaying into two charged muons
$$B^0 -> \mu^++\mu^-$$
(actually this decay is measured in the article at the level of ~3$\sigma$, but I guess new measurements must have increased this significance). My question is what is the physical reason of a much lower probability of finding this decay rather than any other decay from charged B-mesons (e.g.):
$$B^+ -> \mu^++\nu_\mu$$
 A: My first instinct was to exhort you to take a detailed course on the SM where all the precise cogs of your cited paper's Fig 1 are detailed,

but then I appreciated you wished to know, crudely, what everyone in the audience of a talk on this automatically does in checking their assertions in the caption... and nod when they are satisfied.... a) and c) are standard parts of a course beaten to death there; but, because it is simplest, let us back-of-the-envelope estimate  the ratio of the magnitudes of the amplitudes of e) versus b), and square it to get the ratio of the rates and branching ratios. 
Veery crudely indeed... (So we ignore CKM suppressions they talk about, where a skip across one generation amounts to a factor of Wolfenstein λ ~ 0.22 in the amp and square of that in the B.R. So e) would compare, in this respect with b) involving a C instead of a u... this is peanuts.) 
Likewise, the helicity suppression of a spinless B decaying into two back-to back left-chiral leptons in its rest frame, like the celebrated a), is in common between b) and e): the heaviest lepton dominates by violating helicity, here the μ, but it does roughly the same in both processes.
So we only need compare relative propagator suppressions. e) compares to b) by having one more W propagator, and one t propagator , other things being comparable. Since the B is ~ 5 GeV, the internal momenta involved are dominated by the ~ 1 GeV region, and let's pretend the masses of the t and the W are ~ 100 GeV (yes, above and below respectively, but their erring departures compensate). Since the common energy scale is 1GeV, drop GeVs in all dimensionally consistent formulas.
The upshot is 
$$
\frac{\hbox{Amp } e)}{\hbox{Amp } b)} \approx \frac {1}{m_W^2} \frac{1}{m_t} \sim 10^{-1-2\cdot 2}\sim 10^{-5}, 
$$
squaring to ~ $10^{-10}$ they almost anticipate shortly after that figure, noting the purely leptonic charged decays are not quite dominant... they are small.  
This is the type of unashamedly crude estimate people smile about in talks, but it fixes everybody's bearings...  Some pros are monstrously good at it, obviously not I.
