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Question: In roulette, the wheel has \(13\) numbers \(0,1,2---12\) marked on equally spread slots. A player be...

In roulette, the wheel has 1313 numbers 0,1,2−−−120,1,2---12 marked on equally spread slots. A player bets Rs.1010 on a given number. He receives Rs.100100 from the organiser of the game, if the ball comes to rest in this slot, otherwise, he gets nothing. If XX denotes the players net gain/loss. If E(X)=−30XE\left( X \right)=-\dfrac{30}{X}, then find XX.

Explanation

Solution

Here the total number is 1313 and favourable outcome is 11. 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 1313 numbers are marked in equal space. And a player bets on only one number for Rs.1010 to the organiser. If the wheel stops at that number chosen by the player then the organiser has to give Rs.100100 to that player. If he loses, he will get no rupees.
So, here a player bets Rs.1010 on a number, if he wins he will get Rs.100100, which means profit of Rs.9090 but if he loses, he suffers a loss of Rs.1010.
So, here total possible outcomes =13=13
(This means total how many no. is possible to which wheel stops.)
Favourable outcomes =1=1
Favourable outcome means the number chosen by the player. So, here the player is choosing only one number. So, favourable outcomes =1=1
Probability of getting profit =113=\dfrac{1}{13}
Probability of getting loss =1−=1- P (getting profit)
=1−113=1-\dfrac{1}{13}
=1213=\dfrac{12}{13}
So, here we describe XX as the random variable which denotes the gain and loss of the player.
So, possible values of XX can be
X=90X=90 ,i.e., he will get profit
X=−10X=-10, i.e., he will get loss
So, given
P(X=90)=113P\left( X=90 \right)=\dfrac{1}{13}
P(X=−10)=1213P\left( X=-10 \right)=\dfrac{12}{13}
Now let us find the mean

Xi{{X}_{i}}Pi{{P}_{i}}XiPi{{X}_{i}}{{P}_{i}}
9090113\dfrac{1}{13}9013\dfrac{90}{13}
−10-101213\dfrac{12}{13}−12013\dfrac{-120}{13}

Mean =E(X)=∑PiXi=9013−12013=−3013=E\left( X \right)=\sum\limits_{{}}^{{}}{{{P}_{i}}{{X}_{i}}=\dfrac{90}{13}-\dfrac{120}{13}=-\dfrac{30}{13}} (1)\left( 1 \right)
Now in the question, it is given that
E(X)=−30XE\left( X \right)=-\dfrac{30}{X}
And we found that
E(X)=−3013E\left( X \right)=-\dfrac{30}{13}
So, on comparing both the values we can write
X=13X=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(X)=∑PiXi.E\left( X \right)=\sum\limits_{{}}^{{}}{{{P}_{i}}{{X}_{i}}}.