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2 answers:
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Answer:
so at the long run we can conclude that the best option is :
A) win 0.20 cents per play
Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
Let X the random variable who represent the ampunt of money win/loss at the game defined.
The probability of loss $3.00 for this game is 0.2 and the probability of win is 1-0.2=0.8 and you will recieve $1.00 if you win. The expected value is given by:
And for this case if we replace we got:
so at the long run we can conclude that the best option is :
A) win 0.20 cents per play