<s>[INST] M tosses $7$ fair coins and has $M$ heads. $A$ tosses $6$ fair coins and has $A$ heads. Find probability $P(M>A).$ $M$ tosses $7$ fair coins and has M heads. A tosses $6$ fair coins and has $A$ heads. Find probability $P(M>A)$.
I suppose that both distributions are binominal, but I don't know what to do next.
 [/INST] The values $6$ and $7$ are sufficiently small that you could calculate this by hand.
The probabilty that $A$ gets zero heads is $ \frac{1}{64}$ and $M$ get one or more heads is $\frac{127}{128}$.
The probabilty that $A$ gets one head is $ \frac{6}{64}$ and $M$ get two or more heads is $\frac{120}{128}$.
...
The probabilty that $A$ gets six heads is $ \frac{1}{64}$ and $M$ gets seven heads is $\frac{1}{128}$.
We have
\begin{eqnarray*}
P= \frac{1}{64} \frac{127}{128}+\frac{6}{64} \frac{120}{128}+\frac{15}{64} \frac{99}{128}+\frac{20}{64} \frac{64}{128}+\frac{15}{64} \frac{29}{128}+\frac{6}{64} \frac{8}{128}+\frac{1}{64} \frac{1}{128}=\frac{4096}{8192}
\end{eqnarray*}
So the probability is $\color{red}{\frac{1}{2}}$.
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