1. Tardigrade
  2. Question
  3. Mathematics
  4. Let Bn denotes the event that n fair dice are rolled once with P(Bn)=(1/2n), n ∈ N. e.g. P(B1)=(1/2), P(B2)=(1/22), P(B3)=(1/23), ldots ldots ldots and P(Bn)=(1/2n) Hence B 1, B 2, B 3, ldots ldots ldots B n are pairwise mutually exclusive and exhaustive events as n arrow ∞. The event A occurs with atleast one of the event B1, B2, ldots ldots ., Bn and denotes that the sum of the numbers appearing on the dice is S. Probability that greatest number on the dice is 4 if three dice are known to have been rolled, is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 $S$.
Probability that greatest number on the dice is 4 if three dice are known to have been rolled, is

Probability - Part 2

Solution:

$n(S)=6 \cdot 6 \cdot 6=6^3=216$
Now greatest number is 4 , so atleast one of the dice shows up 4 .
$\therefore n ( A )=4^3-3^3=37$
Hence $P(A)=\frac{37}{216}$