Answer:
0.2% or 0.002
Step-by-step explanation:
If X is a random variable then expected value of X is represented by E(X) which is given as:
[tex]E(X)=\sum\limits^{n}_{i=1} {x_{i} } \, p_{i}[/tex]
where [tex]x_{i}[/tex] means observation and [tex]p_{i}[/tex] is the probability of the observation.
Let if probability of winning is p
then probability of losing is 1-p
If the bet (of $1000) is win, then net gain is $9900
If bet is lose, then it means loss and loss in this case will be -100.
Thus, the expected value of the game is = E(X)=$100.
According to above discussion we get the following as probability distribution:
For probability=p, the value of [tex]x_{i}[/tex] is 9900
and for probability= 1-p, the value of[tex]x_{i}[/tex] is -100
[tex]E(X)=\sum\limits^{n}_{i=1} {x_{i} } \, p_{i}=9900(p)+(-100)(1-p)[/tex]
[tex]100=900p-100+100p[/tex]
[tex]200=1000p[/tex]
[tex]p=\frac{200}{1000}[/tex]
[tex]p=0.002 or 0.2%[/tex]
Probability of winning is 0.002 or 0.2%
Although probability of winning is very low but the expected value is positive that is why we should play the game.
If we assume this game to be an experiment, the random variable X will have outcomes as winning amounts, and its expected value thus represents the expected average winnings of the game.
The expected value of this game is positive so I would play this game.
