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# In a game, a man wins 100 rupees if he gets 5 or 6 on a throw of a fair die and loses 50 rupees for getting any other number on the die. If he decides to throw the die either is: till he gets a five or a six or to a maximum of three throws, then his expected gain/ loss (in rupees)A) $\dfrac{{400}}{3}$ gainB) $\dfrac{{400}}{3}$ lossC) 0D) $\dfrac{{400}}{9}$ loss

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Hint:In the question, we are interested in the probability of getting 5 or 6 and also with the amount of gain/loss which is associated with the probability of getting 5 or 6. This type of question is solved by Random Variable and its Probability Distribution.
A random variable is a real valued function whose domain is the sample space of a random experiment.
For example, let us consider the experiment of tossing a coin two times in succession.
The sample space of the experiment is $S = \left\{ {{\text{HH, HT, TH, TT}}} \right\}$.
If X denotes the number of heads obtained, then X is a random variable and for each outcome, its value is as given below:
${\text{X}}\left( {{\text{HH}}} \right) = 2,{\text{ X}}\left( {{\text{HT}}} \right) = 1,{\text{ X}}\left( {{\text{TH}}} \right) = 1,{\text{ X}}\left( {{\text{TT}}} \right) = 0.$
The description (or table) giving the values of the random variables along with the corresponding probabilities is called the probability distribution of the random variable X.
Probability distribution for above example:
 X 0 1 2 $P\left( X \right)$ $\dfrac{1}{4}$ $\dfrac{1}{2}$ $\dfrac{1}{4}$

In the given question we have to find the expected value, the mean or expectation of a random variable X , denoted by $E\left( X \right)$, is the sum of the products of all possible values of X by their respective probabilities.
$E\left( X \right) = 0 \times \dfrac{1}{4} + 1 \times \dfrac{1}{2} + 2 \times \dfrac{1}{4}$

Complete step-by-step answer:
Step 1: calculate the probability of success and failure of the experiment.
Let X be a random variable i.e. the amount of rupees gain/loss in the game.
Let success, p is getting 5 or 6 on a throw.
Probability (5 or 6) $= \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6}$
$\Rightarrow$ Success,$p = \dfrac{1}{3}$
Let failure, q is getting other than 5 or 6 on a throw.
Probability (other than 5 or 6) $= \dfrac{4}{6}$
$\Rightarrow$ Failure,$q = \dfrac{2}{3}$
Maximum numbers of throws = 3.
Step 2: Consider the following cases:
Case 1: Getting 5 or 6 in first throw.
$\Rightarrow$person win amount of 100 rupees,
$\because $$X = 100, P\left( {X = 100} \right) = {\text{ Success in first throw}} \\ {\text{ }} = p = \dfrac{1}{3} \\ Case 2: Getting 5 or 6 in the second throw, that means in the first throw there was a number other than 5 or 6. \Rightarrow person lose amount of 50 rupees in first throw and then win 100 rupees in second throw \because$$X = - 50 + 100 = 50$,
$P\left( {X = 50} \right) = {\text{ Failure in first throw}} \times {\text{Success in second throw}} \\ {\text{ }} = q \times p \\ {\text{ }} = \dfrac{2}{3} \times \dfrac{1}{3} = \dfrac{2}{9} \\$
Case 3: Getting 5 or 6 in the third throw, that means in the first and second throw there were numbers other than 5 or 6.
$\Rightarrow$person lose twice amount of 50 rupees in first and second throw and then win 100 rupees in third throw
$\because $$X = - 50 - 50 + 100 = 0, P\left( {X = 0} \right) = {\text{ Failure in first throw}} \times {\text{Failure in second throw}} \times {\text{Success in third throw}} \\ {\text{ }} = q \times q \times p \\ {\text{ }} = \dfrac{2}{3} \times \dfrac{2}{3} \times \dfrac{1}{3} = \dfrac{4}{{27}} \\ Case 4: Not getting 5 or 6 consecutively in first, second and third throws. \Rightarrow person lose three times the amount of 50 rupees in first, second and third throw \because$$X = - 50 - 50 - 50 = - 150$,
$P\left( {X = - 150} \right) = {\text{ Failure in first throw}} \times {\text{Failure in second throw}} \times {\text{Failure in third throw}} \\ {\text{ }} = q \times q \times q \\ {\text{ }} = \dfrac{2}{3} \times \dfrac{2}{3} \times \dfrac{2}{3} = \dfrac{8}{{27}} \\$
Thus the Probability Distribution of the given question.

 $X{\text{ or }}\mathop x\nolimits_i$ 100 50 0 -150 $P\left( X \right){\text{ or }}\mathop p\nolimits_i$ $\dfrac{1}{3}$ $\dfrac{2}{9}$ $\dfrac{4}{{27}}$ $\dfrac{8}{{27}}$

Expected value, $E\left( X \right) = \sum\limits_{i = 1}^n {\mathop x\nolimits_i } \mathop p\nolimits_i$
$= 100 \times \dfrac{1}{3} + 50 \times \dfrac{2}{9} + 0 \times \dfrac{4}{{27}} + \left( { - 150} \right) \times \dfrac{8}{{27}} \\ = \dfrac{{100}}{3} + \dfrac{{100}}{9} - \dfrac{{1200}}{{27}} \\ = \dfrac{{900 + 300 - 1200}}{{27}} \\ = \dfrac{0}{{27}} \\ = 0 \\$
Final answer: The expected value of gain/loss is 0. Thus, option (C) is correct.

Note:The sum of probabilities of random distribution is 1
i.e. $\sum\limits_{i = 1}^n {\mathop p\nolimits_i } = 1$
verifying:
The probability distribution table of above question:

 $X{\text{ or }}\mathop x\nolimits_i$ 100 50 0 -150 $P\left( X \right){\text{ or }}\mathop p\nolimits_i$ $\dfrac{1}{3}$ $\dfrac{2}{9}$ $\dfrac{4}{{27}}$ $\dfrac{8}{{27}}$

$\sum\limits_{i = 1}^n {\mathop p\nolimits_i } = \dfrac{1}{3} + \dfrac{2}{9} + \dfrac{4}{{27}} + \dfrac{8}{{27}}$
$\Rightarrow \dfrac{{9 + 6 + 4 + 8}}{{27}} \\ \Rightarrow \dfrac{{27}}{{27}} \\ \Rightarrow 1 \\$
If $\sum\limits_{i = 1}^n {\mathop p\nolimits_i } \ne 1$, then there must be some calculation mistake while solving the question.
Another entity may ask further questions is the variance of random variables.
i.e. $Var\left( x \right) = \sum\limits_{i = 1}^n {\mathop x\nolimits_i^2 } p\left( {\mathop x\nolimits_i } \right) - \mathop {\left( {\sum\limits_{i = 1}^n {\mathop x\nolimits_i } p\left( {\mathop x\nolimits_i } \right)} \right)}\nolimits^2$
And standard deviation of random variables., $\mathop \sigma \nolimits_x = \sqrt {Var\left( x \right)}$.