Question & Answer
QUESTION

Ram and Shyam throw with one dice for a prize of Rs.88 which is to be won by the player who throws 1 first. If Ram starts, then mathematical expectation for Shyam is:
(a) Rs.32
(b) Rs.40
(c) Rs.48
(d) None of these

ANSWER Verified Verified
Hint: In this question, we need to first find the possibilities for shyam to win then from that we get the probability of his winning. Then by multiplying this with the prize we get the mathematical expectation for shyam.

Complete step-by-step solution:

PROBABILITY: If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening or occurrence of A, denoted by P(A), is given by
\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\]
Total number of possible outcomes when we throw a dice are 1, 2, 3, 4, 5, 6.
 Number of favourable outcomes to get 1 is 1
Number of favourable outcomes for not getting 1 are 5.
Let us assume that the probability of getting 1 as \[P\left( A \right)\] and the probability of not getting 1 as \[P\left( B \right)\]
 Now, from the above conditions we get,
\[\begin{align}
  & P\left( A \right)=\dfrac{1}{6} \\
 & P\left( B \right)=\dfrac{5}{6} \\
\end{align}\]
Now, for shyam to win the only possibility is ram should lose because he starts first.
Shyam wins if he throws 1 and getting any other number other than 1 is considered as lost.
From the above condition we get,
Let us assume the probability of shyam winning as \[P\left( C \right)\]
Now, we can write the probability of shyam winning as:
\[\Rightarrow P\left( C \right)=\dfrac{5}{6}\times \dfrac{1}{6}+\dfrac{5}{6}\times \dfrac{5}{6}\times \dfrac{5}{6}\times \dfrac{1}{6}+.....\]
Now, by taking the common terms out we get,
\[\Rightarrow P\left( C \right)=\dfrac{5}{36}\left[ 1+\dfrac{25}{36}+{{\left( \dfrac{25}{36} \right)}^{2}}+.... \right]\]
As we already know that the sum of an infinite GP series can be written as
\[\dfrac{a}{1-r}\]
Where, a is the first term and r is the common ratio.
Now, by substituting the respective values in the above formula we get,
\[\Rightarrow P\left( C \right)=\dfrac{5}{36}\left[ \dfrac{1}{1-\dfrac{25}{36}} \right]\]
Now, this can be further simplified as
\[\Rightarrow P\left( C \right)=\dfrac{5}{36}\times \dfrac{36}{11}\]
\[\therefore P\left( C \right)=\dfrac{5}{11}\]
Thus, the mathematical expectation for shyam will be:
\[\Rightarrow \dfrac{5}{11}\times 88\]
\[\begin{align}
  & \Rightarrow 5\times 8 \\
 & \Rightarrow Rs.40 \\
\end{align}\]
Hence, the correct option is (b).

Note: Instead of considering two different events for the occurrence and non occurrence of one we can directly calculate the probability of getting one and then subtracting this from 1 gives the probability of not getting one. Both the methods give the same result.It is important to note that there are several possibilities for shyam to win because as ram starts first he needs to lose then shyam may win in the first attempt or may not. So, when he does not win again the chance goes to ram where he needs to lose again then shyam gets the chance. This time also shyam may win or may not if not then the process continues. That is why we considered it as the sum of an infinite GP series.