Question

Let $B_n$ denotes the event that $n$ fair dice are rolled once with $P\left(B_n\right)=\frac{1}{2^n},n\in N$.
e.g. $P\left(B_1\right)=\frac{1}{2},P\left(B_2\right)=\frac{1}{2^2},P\left(B_3\right)=\frac{1}{2^3},\dots \dots \dots$ and $P\left(B_n\right)=\frac{1}{2^n}$
Hence $B_1,B_2,B_3,\dots \dots \dots B_n$ are pairwise mutually exclusive and exhaustive events as $n\to \infty$. The event $A$ occurs with atleast one of the event $B_1,B_2,\dots \dots .,B_n$ and denotes that the sum of the numbers appearing on the dice is $S$.
If even number of dice has been rolled, the probability that $S=4$, is

• A. very closed to $\frac{1}{2}$
• B. very closed to $\frac{1}{4}$
• C. very closed to $\frac{1}{8}$
• D. very closed to $\frac{1}{16}$

Answer 1

Answer: (D)

$P\left(A/B_E\right)=\frac{P\left(A\cap B_E\right)}{P\left(B_E\right)}\text{, where }A=\text{ sum is }4.$
$\text{ And }P\left(B_E\right)=P\left(B_2\right)+P\left(B_4\right)+\dots \dots \dots \dots ...$
$=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dots \dots =\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{4}\cdot \frac{4}{3}=\frac{1}{3}$

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