solving this probability paragraph

Question

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.

Answer 1

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.

Answer 2

probability that greatest number on the dice is 4 if three dice are known to have been rolled is $37/216$

This is about a conditional probability $|B_3$. We have $P=P’-P’’$, where $P’=\left(\frac 46\right)=\frac {64}{216}$ is a probability that the greatest number on a dice is at most $4$ and $P’’=\left(\frac 36\right)=\frac {27}{216}$ is a probability that the greatest number on a dice is at most $3$.

If even number of dice has been rolled,then show that probability that $S=4$ is very close to $1/16$

I tried two interpretations for an event $A$ which is $S=4$, but obtained the following answers.

If $A$ means that at least one thrown dice has $4$ then we have

$$P=P\left(A{\Huge|}\bigcup_{k=1}^\infty B_{2k}\right)= \frac{1}{P\left(\bigcup_{k=1}^\infty B_{2k}\right)}\sum_{k=1}^\infty P(A|B_{2k})P(B_{2k}) =$$ $$\frac{1}{\sum_{k=1}^\infty \frac 1{2^{2k}}}\sum_{k=1}^\infty \left(1-\left(\frac 56\right)^{2k}\right)\frac 1{2^{2k}}= 1-\frac{\sum_{k=1}^\infty\left(\frac{5}{12}\right)^{2k}}{\sum_{k=1}^\infty \frac 1{2^{2k}}}=$$ $$1-\frac{\frac{\left(\frac{5}{12}\right)^2}{1-\left(\frac{5}{12}\right)^2}}{\frac {\frac 1{2^2}}{1-{\frac 1{2^2}}} }= 1-\frac{\frac{1}{\left(\frac{12}{5}\right)^2-1}} {\frac 1{\frac {2^2}1-1}}= 1-\frac{2^2-1}{{\left(\frac{12}{5}\right)^2-1}}=$$ $$1-\frac{3\cdot 5^2}{{12^2-5^2}}=1-\frac{75}{119}=\frac{44}{119}.$$

If $A$ means that all thrown dices have $4$ then we have

$$P=P\left(A{\Huge|}\bigcup_{k=1}^\infty B_{2k}\right)= \frac{1}{P\left(\bigcup_{k=1}^\infty B_{2k}\right)}\sum_{k=1}^\infty P(A|B_{2k})P(B_{2k}) =$$ $$\frac{1}{\sum_{k=1}^\infty \frac 1{2^{2k}}}\sum_{k=1}^\infty \frac 1{6^{2k}}\cdot\frac 1{2^{2k}}= \frac{\frac 1{12^2} \cdot\frac {1}{1-\frac 1{12^2}}}{\frac 1{2^2} \cdot\frac {1}{1-\frac 1{2^2}} }= \frac {\frac{1}{12^2-1}}{\frac{1}{2^2-1}}=\frac 3{143}.$$

if $S=3$, then prove that $P(B_2/S)=24/169$

$P(B_2|A)=\frac{P(A\cap B_2)}{P(A)}$, but because of the above there is a problem how to interpret $A$.

Answer 3

An alternative answer that only aims at simplicity and tries to explain the results in an intuitive manner (the question already has very good answers).

Let $n$ be the number of rolled dice. The total probability that $n$ is even is

$$P_{even}=1/4+1/16+1/64...=\sum(1/4)^i=1/3$$

The probability to roll two dice is $1/4$, which accounts for $(1/4)/(1/3)=3/4$ of $P_{even}$ (this is an intuitive way to get the same result obtained using the standard formula of conditional probability). With $n=2$, among $36$ possibile outcomes, the only cases giving a total of $4$ are $[3,1]$, $[2,2]$,$[1,3]$, corresponding to $3/36=1/12$ probability. So, given that $n$ is even, we have $ 3/4 \cdot 1/12=1/16$ probability to get $4$ rolling two dice.

Similarly, the probability to roll four dice is $1/16$, which accounts for $(1/16)/(1/3)=3/16$ of $P_{even}$. With $n=4$, among $6^4=1296$ possibile outcomes, the only case giving a total of $4$ is $[1,1,1,1]$, corresponding to $1/1296$ probability. So, given that $n$ is even, we have $ 3/16 \cdot 1/1296=1/6912$ probability to get $4$ throwing four dice.

Since there are no other ways to get $4$ using an even number of dice, we conclude that the answer to the first question is

$$P(S=4)|\text{(even n)}=\frac{1}{16}+\frac{1}{6912}$$

---

For the second question, the probability $P$ that the greatest number on the dice is 4 for $n=3$ can be determined as follows. All dice have to give values not higher than $4$. If the first die gives $4$, then it is sufficient that the other two dice have values $\leq4$: this occurs with probability $1/6\cdot4/6\cdot4/6=16/216$. If the first die does not give $4$, but this occurs for the second die, then it is sufficient that the third one has a value $\leq4$. This occurs with probability $3/6\cdot1/6\cdot4/6=12/216$. Lastly, if none of the first two dice give $4$, then this has to occur for the third die. This occurs with probability $3/6\cdot3/6\cdot1/6=9/216$. So the answer to the second question is

$$P=\frac{16}{216}+\frac{12}{216} +\frac{9}{216}=\frac{37}{216}$$

---

For the third question, we can again try to solve it in an intuitive fashion. The result $S=3$ can be achieved only in three cases: a value of $3$ with a single die, values of $[2,1]$ or $[1,2]$ rolling two dice, and values $[1,1,1]$ rolling three dice. Reminding that the probability of rolling $n$ dice is $1/2^n$, the first case occurs with probability $1/2\cdot1/6=1/12$, the second one with probability $1/4\cdot (1/6)^2\cdot 2=1/72$, and the third one with probability $1/8\cdot (1/6)^3=1/1728$. Then the total probability to get $S=3$ is

$$P(S=3)=\frac{1}{12}+\frac{1}{72} +\frac{1}{1728}=\frac{169}{1728}$$

Of this total probability, the proportion accounted by the case in which two dice are rolled is

$$P(n=2)|(S=3)=\frac{1/72}{169/1728}=\frac{24}{169}$$