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Question: A person throws two fair dice. He wins Rs.15 for throwing a doublet (same numbers on the two dice), ...

A person throws two fair dice. He wins Rs.15 for throwing a doublet (same numbers on the two dice), wins Rs.12 when the throw results in the sum of 9, and loses Rs.6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is A.$2$ gain
B. 12\dfrac{1}{2}loss C.$\dfrac{1}{4}$ loss
D. 12\dfrac{1}{2} gain$$$$

Explanation

Solution

First find out the size of the sample space and then the probability of showing a doublet, the probability of showing a sum of 9. Use the formula of expectation for two events E=AP(X=a)+BP(X=b)E=AP\left( X=a \right)+BP\left( X=b \right) where X=a,X=bX=a,X=b are events and A,B are corresponding values. $$$$

Complete step by step answer:
We know from definition of probability that if there n(A)n\left( A \right) number of ways of event occurring and n(S)n\left( S \right) is the size of the sample space then the probability of the event AA happening is n(A)n(S)\dfrac{n\left( A \right)}{n\left( S \right)}. Mutually exclusive events are events which do not occur at the same time. It means if there are 2 only possible events say $A$ and $\text{B}$ if mutually exclusive then $$P\left( A\bigcap B \right)=0,P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)=1$$ The possible number of outcomes of throwing one fair die is 6. So the number possible outcomes of throwing two fair die one after another or at once by fundamental principle of counting is $6\times 6=36$. So the size of the sample space is 36.
The person loses Rs.15 if for throwing a doublet ( if both die show the same number). The possible cases are (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\left( 1,1 \right),\left( 2,2 \right),\left( 3,3 \right),\left( 4,4 \right),\left( 5,5 \right),\left( 6,6 \right). So the number of such cases are 6 and the probability of a doublet thrown is 636=16\dfrac{6}{36}=\dfrac{1}{6} and the expected gain from it is 15×1615\times \dfrac{1}{6}.$$$$

The person wins Rs.12 if for throwing a sum of 9. The possible cases are (3,6),(6,3),(4,5),(5,4)\left( 3,6 \right),\left( 6,3 \right),\left( 4,5 \right),\left( 5,4 \right). So the number of such cases are are 4 and the probability of a sum of 9 being thrown is 436=19\dfrac{4}{36}=\dfrac{1}{9} and the expected gain from it is 12×1912\times \dfrac{1}{9}. The person loses Rs.6 if for any other outcomes.. The number of possible outcomes that is not doublet or a sum of 9 is $36-6-4=26$ and the probability of such outcomes is $\dfrac{26}{36}$ and the expected loss from it is $6\times \dfrac{26}{36}$.
As the three occurrence of the given three possible outcomes are mutually exclusive events, the total expected gain or loss in rupees is

P(Loss/Gain)=15×636+12×19−6×2636=−12P\left( \text{Loss/Gain} \right)=15\times \dfrac{6}{36}+12\times \dfrac{1}{9}-6\times \dfrac{26}{36}=\dfrac{-1}{2}

So, the correct answer is “Option B”.

Note: We can also find the expectation of nn number of events X=a1,X=a2,...,X=anX={{a}_{1}},X={{a}_{2}},...,X={{a}_{n}} with respective probabilities P(X=a)1,P(X=a2),...,P(X=an)P{{\left( X=a \right)}_{1}},P\left( X={{a}_{2}} \right),...,P\left( X={{a}_{n}} \right) and respective value of outcomes A1,A2,...,An{{A}_{1}},{{A}_{2}},...,{{A}_{n}} as E(X)E\left( X \right) which is given by
E(X)=P(X=a1)A1+P(X=a2)A2+...+P(X=an)AnE\left( X \right)=P\left( X={{a}_{1}} \right){{A}_{1}}+P\left( X={{a}_{2}} \right){{A}_{2}}+...+P\left( X={{a}_{n}} \right){{A}_{n}}. The problem can be framed to test independence or inclusiveness of events.