solving this probability paragraph Let $B_n$ denotes the event that n  fair dice are
rolled once with $P(B_n)=1/2^n$ where n is a natural number.
Hence $B_1,B_2,B_3,..B_n$are pairwise mutually exclusive events as n approaches infinity.
The event A occurs with atleast one of the event
$B_1,B_2,B_3,..B_n$ and denotes that the numbers appearing on the dice is S
If even number of dice has been rolled,then show that probability that $S=4$ is very close to $1/16$
next show that probability that greatest number on the dice is 4 if three dice are known to have been rolled is $37/216$
Finally,if $S=3$, then prove that
$P(B_2/S)=24/169$
my approach :
well I tried using the conditional probability formula in part one and baye's theorem in the final one but I am unable to get to the correct answer.kindly help me out,all help is greatly appreciated.


probability that greatest number on the dice is 4 if three dice are known to have been rolled is $37/216$

This is about a conditional probability $|B_3$. We have $P=P’-P’’$, where $P’=\left(\frac 46\right)=\frac {64}{216}$ is a probability that the greatest number on a dice is at most $4$ and $P’’=\left(\frac 36\right)=\frac {27}{216}$ is a probability that the greatest number on a dice is at most $3$.

If even number of dice has been rolled,then show that probability that $S=4$ is very close to $1/16$

I tried two interpretations for an event $A$ which is $S=4$, but obtained the following answers.
If $A$ means that at least one thrown dice has $4$ then we have
$$P=P\left(A{\Huge|}\bigcup_{k=1}^\infty B_{2k}\right)=
\frac{1}{P\left(\bigcup_{k=1}^\infty B_{2k}\right)}\sum_{k=1}^\infty P(A|B_{2k})P(B_{2k}) =$$
$$\frac{1}{\sum_{k=1}^\infty \frac 1{2^{2k}}}\sum_{k=1}^\infty \left(1-\left(\frac 56\right)^{2k}\right)\frac 1{2^{2k}}=
1-\frac{\sum_{k=1}^\infty\left(\frac{5}{12}\right)^{2k}}{\sum_{k=1}^\infty \frac 1{2^{2k}}}=$$
$$1-\frac{\frac{\left(\frac{5}{12}\right)^2}{1-\left(\frac{5}{12}\right)^2}}{\frac {\frac 1{2^2}}{1-{\frac 1{2^2}}} }=
1-\frac{\frac{1}{\left(\frac{12}{5}\right)^2-1}} {\frac 1{\frac {2^2}1-1}}=
1-\frac{2^2-1}{{\left(\frac{12}{5}\right)^2-1}}=$$ $$1-\frac{3\cdot 5^2}{{12^2-5^2}}=1-\frac{75}{119}=\frac{44}{119}.$$
If $A$ means that all thrown dices have $4$ then we have
$$P=P\left(A{\Huge|}\bigcup_{k=1}^\infty B_{2k}\right)=
\frac{1}{P\left(\bigcup_{k=1}^\infty B_{2k}\right)}\sum_{k=1}^\infty P(A|B_{2k})P(B_{2k}) =$$
$$\frac{1}{\sum_{k=1}^\infty \frac 1{2^{2k}}}\sum_{k=1}^\infty \frac 1{6^{2k}}\cdot\frac 1{2^{2k}}=
\frac{\frac 1{12^2} \cdot\frac {1}{1-\frac 1{12^2}}}{\frac 1{2^2} \cdot\frac {1}{1-\frac 1{2^2}} }=
\frac {\frac{1}{12^2-1}}{\frac{1}{2^2-1}}=\frac 3{143}.$$

if $S=3$, then prove that $P(B_2/S)=24/169$

$P(B_2|A)=\frac{P(A\cap B_2)}{P(A)}$,  but because of the above there is a problem how to interpret $A$.
