Question

In roulette, the wheel has $13$ numbers $0,1,2---12$ marked on equally spread slots. A player bets Rs.$10$ on a given number. He receives Rs.$100$ from the organiser of the game, if the ball comes to rest in this slot, otherwise, he gets nothing. If $X$ denotes the players net gain/loss. If $E\left( X \right)=-\dfrac{30}{X}$, then find $X$.

Answer 1

Hint: Here the total number is $13$ and favourable outcome is $1$. We can find the probability of getting profit as well as the probability of getting lost.

Complete step by step solution:
Basically, here it is given a wheel in which total $13$ numbers are marked in equal space. And a player bets on only one number for Rs.$10$ to the organiser. If the wheel stops at that number chosen by the player then the organiser has to give Rs.$100$ to that player. If he loses, he will get no rupees.
So, here a player bets Rs.$10$ on a number, if he wins he will get Rs.$100$, which means profit of Rs.$90$ but if he loses, he suffers a loss of Rs.$10$.
So, here total possible outcomes $=13$
(This means total how many no. is possible to which wheel stops.)
Favourable outcomes $=1$
Favourable outcome means the number chosen by the player. So, here the player is choosing only one number. So, favourable outcomes $=1$
Probability of getting profit $=\dfrac{1}{13}$
Probability of getting loss $=1-$ P (getting profit)
$=1-\dfrac{1}{13}$
$=\dfrac{12}{13}$
So, here we describe $X$ as the random variable which denotes the gain and loss of the player.
So, possible values of $X$ can be
$X=90$ ,i.e., he will get profit
$X=-10$, i.e., he will get loss
So, given
$P\left( X=90 \right)=\dfrac{1}{13}$
$P\left( X=-10 \right)=\dfrac{12}{13}$
Now let us find the mean

${{X}_{i}}$	${{P}_{i}}$	${{X}_{i}}{{P}_{i}}$
$90$	$\dfrac{1}{13}$	$\dfrac{90}{13}$
$-10$	$\dfrac{12}{13}$	$\dfrac{-120}{13}$

Mean $=E\left( X \right)=\sum\limits_{{}}^{{}}{{{P}_{i}}{{X}_{i}}=\dfrac{90}{13}-\dfrac{120}{13}=-\dfrac{30}{13}}$ \[\left( 1 \right)\]
Now in the question, it is given that
$E\left( X \right)=-\dfrac{30}{X}$
And we found that
$E\left( X \right)=-\dfrac{30}{13}$
So, on comparing both the values we can write
$X=13$

Note: The profit and loss statement is a financial statement that summarizes the revenues, costs, and expenses incurred during a specific period, usually a fiscal quarter or year. These records provide information about a company’s ability or inability to generate profit by increasing revenue, reducing costs, or both.
A player can get profit or loss. So, mean can be calculated by probability distribution method, i.e. $E\left( X \right)=\sum\limits_{{}}^{{}}{{{P}_{i}}{{X}_{i}}}.$