Probability of occurence of alphabets A circular disc with 3 sectors marked as alphabet O,A,B has
circumference l. The alphabet coming infront of marker is noted
down aftr each spin (as shown in diagram). Assume boundary
of sectors do not come in front of marker. It is spun 6 times around
its centre and resulting alphabet which comes infront of marker
are noted down in order. lf the probability that; the word. BAOBAA
to be formed is maximised, than (where x,y,z respectlvely denotes
the probability of occurrence of alphabet O,A,B)
W have to maximixe zyxzyy
So I used AM>GM
That is $xy^3z^2 < (27)(4)/(6^6)$
But how how to proced
 A: It sounds like you're almost there. We want to choose $x,y,z\geq 0$ with $x+y+z=1$ such that $xy^3z^2$ is as large as possible. Applying AM-GM to the numbers $x,\frac y3,\frac y3,\frac y3,\frac z2,\frac z2$, we get $$\sqrt[6]{\frac{xy^3z^2}{108}}\leq\frac 16,$$with equality if and only if $x=\frac y3=\frac z2$. Since in that case $x+y+z=6x=1$, the probability is maximised exactly when $x=\frac 16$, $y=\frac 12$ and $z=\frac 13$.
A: *

*$p(a)=1-p(b)-p(c)$, 

*$p(cbacbb)= \frac{p(c)^2 p(a)^3p(a)}{{6 \choose 3}{3 \choose 2}}=\frac{1}{60}p(c)^2 p(b)^3p(a) = \frac{1}{60}p(c)^2 p(b)^3(1-p(c)-p(b))$

*$f(y,z)=60 p(cbacbb) = p(c)^2 p(b)^3(1-p(b)-p(c))  = y^2z^3(1-y-z)$

*$\frac{\partial f(y,z) }{\partial y} = y^2z^2(3-4y-3z)\\
\frac{\partial f(y,z) }{\partial z} = y^3z(2-2y-3z)$
$$\frac{\partial f(y,z) }{\partial y}=\frac{\partial f(x,y) }{\partial z}=0 \iff (y,z)\in \left\{\left(\frac{1}{2},0\right),\left(\frac{1}{2},\frac{1}{3}\right)\right\}$$
If we take $(y,z)=\left(\frac{1}{2},0\right)$, then $p(cbacbb)=0$, so we will skip this case.

*$\frac{\partial^2 f(y,z) }{\partial y^2}=6yz^2(1-2y-z)\\
\frac{\partial^2 f(y,z) }{\partial z^2}=2y^3(1-y-3z)\\
\frac{\partial^2 f(y,z) }{\partial y\partial z}=y^2z(6-8y-9z)$
$$H(y,z)=\left| \begin{array}. 6yz^2(1-2y-z) & y^2z(6-8y-9z) \\
y^2z(6-8y-9z) & 2y^3(1-y-3z)\end{array} \right|$$
$\frac{\partial^2 f(y,z) }{\partial y^2}|_{(y,z)=\left(\frac{1}{2},\frac{1}{3}\right)}=-\frac{1}{9}<0\\
H\left(\frac{1}{2},\frac{1}{3}\right)=-\frac{1}{8}\frac{1}{72} <0$
so in point $(y,z)=\left(\frac{1}{2},\frac{1}{3}\right)$ function $f$ have the local maximum.
The probability is them maximized, if $(p(a),p(b),p(c))=\left(\frac{1}{6},\frac{1}{2},\frac{1}{3}\right)$
