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. $P\left(B_1\right)=\frac{1}{2}, P\left(B_2\right)=\frac{1}{2^2}, P\left(B_3\right)=\frac{1}{2^3}, \ldots \ldots \ldots$ and $P\left(B_n\right)=\frac{1}{2^n}$
Hence $B _1, B _2, B _3, \ldots \ldots \ldots B _{ n }$ are pairwise mutually exclusive and exhaustive events as $n \rightarrow \infty$. The event $A$ occurs with atleast one of the event $B_1, B_2, \ldots \ldots ., B_n$ 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 . $
$\text { And } P \left( B _{ E }\right)= P \left( B _2\right)+ P \left( B _4\right)+\ldots \ldots \ldots \ldots . . .$
$=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots \ldots=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{4} \cdot \frac{4}{3}=\frac{1}{3}$

$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 $P\left(A / B_E\right)=\frac{433}{16 \cdot 6^4} \cdot 3=\frac{433}{32 \cdot 216} \cong \frac{1}{16}$