Question

Let $B_n$ denote the event that $n$ fair dice are rolled once with probability $P\left(B_n\right)=\frac{1}{2^n},n\in N.$
e.g. $P\left(B_1\right)=\frac{1}{2},P\left(B_2\right)=\frac{1}{2^2},\dots ,P\left(B_n\right)=\frac{1}{2^n}.$ Hence, $B_1,B_2,B_3,\dots ,B_n$ are pairwise mutually exclusive and exhaustive events as $n\to \infty .$
The event $A$ occurs with one of the events $B_1,B_2,\dots ,B_n.$ Let $S$ denote the sum of the numbers appearing on the dice.

• A. If even number of dice has been rolled, the probability that $S$ equals $4,$ is very close to $\frac{1}{2}$
• B. If even number of dice has been rolled, the probability that $S$ equals $4,$ is very close to $\frac{1}{16}$
• C. Probability that the greatest number on the dice is $4$ if three dice are known to have been rolled, is $\frac{37}{216}$
  
• D. If $S$ equals $3,$ then $P\left(B_2|S\right)$ has the value equal to $\frac{24}{169}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B If even number of dice has been rolled, the probability that $S$ equals $4,$ is very close to $\frac{1}{16}$
C Probability that the greatest number on the dice is $4$ if three dice are known to have been rolled, is $\frac{37}{216}$

D If $S$ equals $3,$ then $P\left(B_2|S\right)$ has the value equal to $\frac{24}{169}$

$\to$ Let $B_E$ be the event that even number of dice has been rolled.
$P\left(\right(S=4\left)|B_E\right)=\frac{P\left(\right(S=4)\cap B_E)}{P\left(B_E\right)}$

$P\left(B_E\right)=P\left(B_2\right)+P\left(B_4\right)+\cdots$
$=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\cdots \infty$
$=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}$

$P\left(\right(S=4)\cap B_E)=P\left(\right(S=4)\cap B_2)+P\left(\right(S=4)\cap B_4)$
$=P\left(B_2\right)P\left(\right(S=4\left)|B_2\right)+P\left(B_4\right)P\left(\right(S=4\left)|B_4\right)$
$=\frac{1}{4}\left(\frac{3}{36}\right)+\frac{1}{16}\left(\frac{1}{6^4}\right)$
$=\frac{3\times 4\times 6^2+1}{16\times 6^4}$
$=\frac{433}{16\cdot 6^4}$
Hence, $P\left(\right(S=4\left)|B_E\right)=\frac{\frac{433}{16\cdot 6^4}}{1/3}$
$=\frac{433}{32\cdot 216}\approx \frac{1}{16}$

$\to$ Three dice are rolled.
Number of possible elements $=6\times 6\times 6=216$
Now greatest number is $4$, so at least one of the dice shows up $4$.
Number of favourable cases $=4^3-3^3=37$
Required probability $=\frac{37}{216}$

$\to$ $S=3$
$P\left(S\right)=P(B_1\cap S)+P(B_2\cap S)+P(B_3\cap S)$
$=\frac{1}{2}\left(\frac{1}{6}\right)+\frac{1}{2^2}\left(\frac{2}{36}\right)+\frac{1}{2^3}\left(\frac{1}{6^3}\right)$
$=\frac{1}{12}+\frac{1}{72}+\frac{1}{\left(8\right)\left(216\right)}$
$=\frac{1}{\left(8\right)\left(216\right)}[144+24+1]=\frac{169}{\left(8\right)\left(216\right)}$

Now $P\left(B_2|S\right)=\frac{P(B_2\cap S)}{P\left(S\right)}$
Hence, $P\left(B_2|S\right)=\frac{1}{72}\times \frac{\left(8\right)\left(216\right)}{169}=\frac{24}{169}$