Questions & Answers

Question

Answers

[a] A is Rs 10

[b] B is Rs 10

[c] A is Rs 20

[d] B is Rs 20

Answer
Verified

Define ${{A}_{i}}$ as the event A wins the ${{i}^{th}}$ toss and ${{B}_{i}}$ as the event B wins the ${{i}^{th}}$ toss.

Let E be the event that A wins the game.

Note that for the ${{i}^{th}}$ toss to take place A and B must all of their i-1 tosses.

Hence, we have

$P\left( {{A}_{i}}|\bigcap\limits_{r=1}^{i-1}{{{A}_{r}}'}\bigcap\limits_{r=1}^{i-1}{{{B}_{r}}'} \right)=0.5$ and $P\left( {{A}_{i}}'|\bigcap\limits_{r=1}^{i-1}{{{A}_{r}}'}\bigcap\limits_{r=1}^{i-1}{{{B}_{r}}'} \right)=0.5$

Similarly, we have

$P\left( {{B}_{i}}|\bigcap\limits_{r=1}^{i}{{{A}_{r}}'}\bigcap\limits_{r=1}^{i-1}{{{B}_{r}}'} \right)=0.5$ and $P\left( {{B}_{i}}'|\bigcap\limits_{r=1}^{i}{{{A}_{r}}'}\bigcap\limits_{r=1}^{i-1}{{{B}_{r}}'} \right)=0.5$

Hence, we have

$P\left( A \right)=P\left( {{A}_{1}} \right)+P\left( {{A}_{1}}'\bigcap {{B}_{1}}'\bigcap {{A}_{2}} \right)+P\left( {{A}_{1}}'\bigcap {{B}_{1}}'\bigcap {{A}_{2}}'\bigcap {{B}_{2}}'\bigcap {{A}_{3}} \right)+\cdots $

Now, we have

$P\left( {{A}_{1}} \right)=\dfrac{1}{2}$(Since the coin is fair and it is 0.5 probable that A guesses correctly)

$\begin{align}

& P\left( {{A}_{1}}'\bigcap {{B}_{1}}'\bigcap {{A}_{2}} \right)=P\left( {{A}_{1}}' \right)\times P\left( {{B}_{1}}'|{{A}_{1}}' \right)\times P\left( {{A}_{2}}|{{B}_{1}}'\bigcap {{A}_{1}}' \right) \\

& =\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{8} \\

\end{align}$

Similarly, we have

$P\left( {{A}_{1}}'\bigcap {{B}_{1}}'\bigcap {{A}_{2}}'\bigcap {{B}_{2}}'\bigcap {{A}_{3}} \right)=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{32}$

Hence, we have

$P\left( A \right)=\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\cdots $ which is an infinite G.P with common ratio of $\dfrac{1}{4}$

We know that the sum of an infinite G.P with first term as a and the common ratio as r, 0 < |r| < 1, is given by ${{S}_{\infty }}=\dfrac{a}{1-r}$

Since $0\ <\ \dfrac{1}{4}\ <\ 1$, we have

$P\left( A \right)=\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{4}}=\dfrac{2}{3}$

Hence probability that A wins is $\dfrac{2}{3}$

Hence the expected amount that A wins is $\dfrac{2}{3}\times 30=20$ and the expected amount that B wins is $\dfrac{1}{3}\times 30=10$. Hence option [b] and [c] are correct.

We can use linearity of the expectation to determine the expected amount that A wins

Let ${{X}_{i}}$ is the amount that A wins in the ${{i}^{th}}$ round.

Hence, we have

$E\left( A \right)=E\left( {{X}_{1}}+{{X}_{2}}+\cdots \infty \right)$

But since the occurrence of ${{X}_{i}}$ is dependent on the event ${{A}_{i-1}}$ and ${{B}_{i-1}}$ and as there is $\dfrac{1}{{{2}^{i-1}}}$ probability that in all the i-1 tosses A loses and probability $\dfrac{1}{{{2}^{i-1}}}$ probability that B losses all the i-1 tosses. Also the probability that A wins the ${{i}^{th}}$ toss is $\dfrac{1}{2}$ and the probability that A loses the ${{i}^{th}}$ toss is $\dfrac{1}{2}$

We have

$\begin{align}

& E\left( {{X}_{i}} \right)=E\left( {{X}_{i}}|{{A}_{i-1}}'\bigcap {{B}_{i-1}}' \right)P\left( {{A}_{i-1}}'\bigcap {{B}_{i-1}}' \right) \\

& =\dfrac{1}{{{2}^{i-1}}}\times \dfrac{1}{{{2}^{i-1}}}\left( 30\times \dfrac{1}{2}+0\times \dfrac{1}{2} \right) \\

\end{align}$

Hence, we have

$E\left( A \right)=15\times \sum\limits_{i=1}^{\infty }{\dfrac{1}{{{2}^{2i-2}}}}$

The term inside the summation is the general term of an infinite G.P with first term 1 and common difference $\dfrac{1}{4}$

Hence, we have

$E\left( A \right)=15\times \dfrac{1}{1-\dfrac{1}{4}}=\dfrac{15\times 4}{3}=20$

Hence option [c] is correct.

Also, we have

$\begin{align}

& E\left( A \right)+E\left( B \right)=30 \\

& \Rightarrow E\left( B \right)=30-20=10 \\

\end{align}$

Hence option [b] is correct.

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