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A coin is tossed repeatedly. A and B call alternatively for winning a prize of Rs 30. One who calls correctly first wins the prize. A starts the call. The expectation of
[a] A is Rs 10
[b] B is Rs 10
[c] A is Rs 20
[d] B is Rs 20

Answer Verified Verified
Hint: Use the fact that A wins the game if A wins if he wins the first toss, or if he wins he loses the first toss, then B also loses the first toss and then A wins the third toss and so on. Use the fact that in each toss the probability of the tosser to win is 0.5 and the probability for the tosser to lose is 0.5. Hence determine the probability of A winning. Use the fact that if p is the probability of A winning then the expected amount that A wins is $Ap$ and the expectation of B is $B\left( 1-p \right)$. Hence verify which of the options is correct.

Complete step-by-step answer:
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.

Note:Alternative Solution:
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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