Respuesta :
Answer:
[tex]E(x)=-0.2101[/tex]
Step-by-step explanation:
The expected value for a discrete variable is calculated as:
[tex]E(x)=x_1*p(x_1)+x_2*p(x_2)[/tex]
Where [tex]x_1[/tex] and [tex]x_2[/tex] are the values that the variable can take and [tex]p(x_1)[/tex] and [tex]p(x_2)[/tex] are their respective probabilities.
So, a player can win 2 dollars or looses 1 dollar, it means that [tex]x_1[/tex] is equal to 2 and [tex]x_2[/tex] is equal to -1.
Then, we need to calculated the probability that the player win 2 dollars and the probability that the player loses 1 dollar.
If there are n identical and independent events with a probability p of success and a probability (1-p) of fail, the probability that a events from the n are success are equal to:
[tex]P(a)=nCa*p^a*(1-p)^{n-a}[/tex]
Where [tex]nCa=\frac{n!}{a!(n-a)!}[/tex]
So, in this case, n is number of times that the player tosses a die and p is the probability to get a 6. n is equal to 6 and p is equal to 1/6.
Therefore, the probability [tex]p(x_1)[/tex] that a player get at least two times number 6, is calculated as:
[tex]p(x_1)=P(x\geq2)=1-P(0)-P(1) \\\\P(0) =6C0*(1/6)^{0}*(5/6)^{6}=0.3349\\P(1)=6C1*(1/6)^{1}*(5/6)^{5}=0.4018\\\\p(x_1)=1-0.3349-0.4018\\p(x_1)=0.2633[/tex]
On the other hand, the probability [tex]p(x_2)[/tex] that a player don't get at least two times number 6, is calculated as:
[tex]p(x_2)=1-p(x_1)=1-0.2633=0.7367[/tex]
Finally, the expected value of the amount that the player wins is:
[tex]E(x)=x_1*p(x_1)+x_2*p(x_2)\\E(x)=2*(0.2633)+ (-1)*0.7367\\E(x)=-0.2101[/tex]
It means that he can expect to loses 0.2101 dollars.
Answer:
Expected value = -0.2101
Step-by-step explanation :
Givena discrete variable the expected value is given as:
E(x) = xp(x) + yp(y)
Where x and y are the values the variables can assume, and p(x) and p(y) are their respective probabilities.
Probability that a player gets 6 at least twice = p(x ≥ 2) = 1 - p(x≤ 2)
= 1 - p(0) - p(1)
But p(0) = 0.3349
p(1) = 0.4018
p(x ≥ 2) = 1 - 0.3349 - 0.4018
= 0.2633
p(y) = 1 - p(x)
= 1 - 0.2633
= 0.7367
Expected value = 2(0.2633) + (-1)(0.7367)
= -0.2101
A loss of $0.2101 is expected.
