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Juli2301 [7.4K]
2 years ago
11

Suppose you play a game in which 5 dice (6-sided) are rolled simultaneously.

Mathematics
1 answer:
goblinko [34]2 years ago
3 0

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 =  E(X) = \sum_{\forall x_i} f(x_i)x_i

Here,   x_i; \: \: i = 1,2, ... , n is its n data values and f(x_i)is the probability of  X = x_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

<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)

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}

The expected value and variance of X are:

E(X) = np\\ Var(X) = np(1-p)

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\\

The probability of loosing $5 equal probability of 0 success.

P(Y=-5)=P(x=0)

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)

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

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

#SPJ1

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