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.
Best Answer
I'm guessing $\ S\ $ is the sum of all the numbers on the dice thrown. If that assumption is correct, then $$ P\left(\bigcup_{n=1}^\infty B_{2n}\right)=\sum_{i=1}^\infty\frac{1}{2^{2i}}=\frac{1}{3}\ , $$ and \begin{align} P\left(\{S=4\}\cap \bigcup_{n=1}^\infty B_{2n}\right)&=P\left(\{S=4\}\cap B_2\right)+P(\{S=4\} \cap B_4 )\\ &=\frac{1}{4}\cdot\frac{1}{12}+\frac{1}{16}\frac{1}{6^4}\\ &=\frac{433}{20736} \end{align} because the sum of the numbers on the dice must exceed $4$ if any other even number of them were to be thrown. Thus \begin{align} P\left(S=4\,\left|\,\bigcup_{n=1}^\infty B_{2n}\right.\right)&=\frac{P\left(\{S=4\}\cap \bigcup_{n=1}^\infty B_{2n}\right)}{P\left(\bigcup_{n=1}^\infty B_{2n}\right)}\\ &=3\cdot \frac{433}{20736}\\ &=\frac{433}{6912}\\ &=\frac{1}{16}+\frac{1}{6912}\ . \end{align} Also \begin{align} P(B_2\cap \{S=3\})&=\frac{1}{4}\cdot\frac{1}{18}\\ p(S=3)&=\sum_{n=1}^3P(B_n\cap\{S=3\})\\ &=\frac{1}{2}\cdot\frac{1}{6}+\frac{1}{4}\cdot\frac{1}{18}+ \frac{1}{8}\cdot\frac{1}{216}\\ &=\frac{169}{1728}\ . \end{align} Therefore \begin{align} P(B_2\,|\,S=3)&=\frac{P(B_2\cap\{ S=3\})}{P(S=3)}\\ &=\frac{1728}{169\cdot4\cdot18}\\ &=\frac{24}{169}\ . \end{align} The remaining part of the question has already been answered by Alex Ravsky.