The expected value of the game in which 5 dice (6-sided) are rolled simultaneously is -0.2094.
<h3>How to find the mean (expectation) and variance of a random variable?</h3>
Supposing that the considered random variable is discrete, we get:
Mean =
![\sum_{\forall x_i} f(x_i)x_i](https://tex.z-dn.net/?f=%5Csum_%7B%5Cforall%20x_i%7D%20f%28x_i%29x_i)
Here,
n is its n data values and
is the probability of ![X = x_i](https://tex.z-dn.net/?f=X%20%3D%20x_i)
Suppose you play a game in which 5 dice (6-sided) are rolled simultaneously.
- If a "4" is rolled, then you win $2 for each "4" showing.
- If all the dice are showing "4", you win $1000.
- If none of the dice are showing "4", then you lose $5.
Let Y is the amount of money player won. The value of X can be,
![Y=2,4,6,8,1000,-5](https://tex.z-dn.net/?f=Y%3D2%2C4%2C6%2C8%2C1000%2C-5)
<h3>How to find that a given condition can be modeled by binomial distribution?</h3>
Binomial distributions consists of n independent Bernoulli trials. Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))
Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as
![X \sim B(n,p)](https://tex.z-dn.net/?f=X%20%5Csim%20B%28n%2Cp%29)
The probability that out of n trials, there'd be x successes is given by
![P(X =x) = \: ^nC_xp^x(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3Dx%29%20%3D%20%5C%3A%20%5EnC_xp%5Ex%281-p%29%5E%7Bn-x%7D)
The expected value and variance of X are:
![E(X) = np\\ Var(X) = np(1-p)](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np%5C%5C%20Var%28X%29%20%3D%20np%281-p%29)
Put the values as 5 trials for each time 4 appears.
![P(X =0) = \: ^5C_1(\dfrac{1}{6})^0(1-\dfrac{1}{6})^{5-0}=0.4018 \:\\P(X =1) = \: ^5C_1(\dfrac{1}{6})^1(1-\dfrac{1}{6})^{5-1}=0.402\\P(X =2) = \: ^5C_2(\dfrac{1}{6})^2(1-\dfrac{1}{6})^{5-2}=0.162\\P(X =3) = \: ^5C_3(\dfrac{1}{6})^3(1-\dfrac{1}{6})^{5-3}=0.032\\P(X =4) = \: ^5C_4(\dfrac{1}{6})^4(1-\dfrac{1}{6})^{5-4}=0.0032\\P(X =5) = \: ^5C_5(\dfrac{1}{6})^5(1-\dfrac{1}{6})^{5-5}=0.00013\\](https://tex.z-dn.net/?f=P%28X%20%3D0%29%20%3D%20%5C%3A%20%5E5C_1%28%5Cdfrac%7B1%7D%7B6%7D%29%5E0%281-%5Cdfrac%7B1%7D%7B6%7D%29%5E%7B5-0%7D%3D0.4018%20%5C%3A%5C%5CP%28X%20%3D1%29%20%3D%20%5C%3A%20%5E5C_1%28%5Cdfrac%7B1%7D%7B6%7D%29%5E1%281-%5Cdfrac%7B1%7D%7B6%7D%29%5E%7B5-1%7D%3D0.402%5C%5CP%28X%20%3D2%29%20%3D%20%5C%3A%20%5E5C_2%28%5Cdfrac%7B1%7D%7B6%7D%29%5E2%281-%5Cdfrac%7B1%7D%7B6%7D%29%5E%7B5-2%7D%3D0.162%5C%5CP%28X%20%3D3%29%20%3D%20%5C%3A%20%5E5C_3%28%5Cdfrac%7B1%7D%7B6%7D%29%5E3%281-%5Cdfrac%7B1%7D%7B6%7D%29%5E%7B5-3%7D%3D0.032%5C%5CP%28X%20%3D4%29%20%3D%20%5C%3A%20%5E5C_4%28%5Cdfrac%7B1%7D%7B6%7D%29%5E4%281-%5Cdfrac%7B1%7D%7B6%7D%29%5E%7B5-4%7D%3D0.0032%5C%5CP%28X%20%3D5%29%20%3D%20%5C%3A%20%5E5C_5%28%5Cdfrac%7B1%7D%7B6%7D%29%5E5%281-%5Cdfrac%7B1%7D%7B6%7D%29%5E%7B5-5%7D%3D0.00013%5C%5C)
The probability of loosing $5 equal probability of 0 success.
![P(Y=-5)=P(x=0)](https://tex.z-dn.net/?f=P%28Y%3D-5%29%3DP%28x%3D0%29)
Similarly, for probability of getting profit are,
![P(Y=2)=P(x=1)\\P(Y=4)=P(x=2)\\P(Y=6)=P(x=3)\\P(Y=8)=P(x=4)\\P(Y=1000)=P(x=5)](https://tex.z-dn.net/?f=P%28Y%3D2%29%3DP%28x%3D1%29%5C%5CP%28Y%3D4%29%3DP%28x%3D2%29%5C%5CP%28Y%3D6%29%3DP%28x%3D3%29%5C%5CP%28Y%3D8%29%3DP%28x%3D4%29%5C%5CP%28Y%3D1000%29%3DP%28x%3D5%29)
Expected value of game,
![E(Y)=\sum y .P(Y=y)\\E(Y)=-5.P(X=0)+2.P(X=1)+4.P(X=2)+6.P(X=3)+8.P(X=4)+1000.P(X=5)\\E(Y)=-0.2094](https://tex.z-dn.net/?f=E%28Y%29%3D%5Csum%20y%20.P%28Y%3Dy%29%5C%5CE%28Y%29%3D-5.P%28X%3D0%29%2B2.P%28X%3D1%29%2B4.P%28X%3D2%29%2B6.P%28X%3D3%29%2B8.P%28X%3D4%29%2B1000.P%28X%3D5%29%5C%5CE%28Y%29%3D-0.2094)
Thus, the expected value of the game in which 5 dice (6-sided) are rolled simultaneously is -0.2094.
Learn more about expectation of a random variable here:
brainly.com/question/4515179
Learn more about binomial distribution here:
brainly.com/question/13609688
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