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You decide to play roulette 4 times, each time betting on red. What is the distribution of X, the number of times you win?

Respuesta :

cutelion918
  1.
If betting on red, 18/38 = 9/19 chance of winning, so 10/19 chance of losing
Chance of winning 4 times: (9/19)^4 = 6561/130321 =~ 5.034%
Chance of winning 3 times: (9/19)^3 * (10/19) * C(4,1) = 29160/130321 =~ 22.3755%
Chance of winning 2 times: (9/19)^2 * (10/19)^2 * C(4,2) = 48600/130321 =~ 37.2925%
Chance of winning 1 time: (9/19) * (10/19)^3 * C(4,3) = 36000/130321 =~ 27.624%
Chance of winning 0 times: (10/39)^4 = 10000/130321 =~ 76.7336%
fichoh

Using the concept of binomial probability, the distribution of X which is the number of winning is ;

  • X ______ 0 ____1 _____ 2___ 3______ 4

  • P(X):_0.0767_0.2762_0.3729_0.2238_0.0503

Given the parmaters :

  • Total Number of slots = 38
  • Number of reds = 18

The probability of winning, p on a single play :

  • p = (red slots / total slots) = 18/38 = 9/19

Number of plays = 4

Number of winnings = X

Using the binomial probability relation :

  • P(x = x) = nCx * p^x * q^(n-x)
  • q = 1 - 9/19 = 10/19
  • n = 4

For X = 0 :

P(x = 0) = 4C0 × (9/19)^0 × (10/19)⁴ = 0.0767

For X = 1 :

P(x = 0) = 4C1 × (9/19)¹ × (10/19)³ = 0.2762

For X = 2 :

P(x = 2) = 4C2 × (9/19)² × (10/19)² = 0.3729

For X = 3 :

P(x = 3) = 4C3 × (9/19)³ × (10/19)¹ = 0.2238

For X = 4 :

P(x = 4) = 4C4 × (9/19)⁴ × (10/19)^0 = 0.0503

Therefore, the distribution of X in the roulette spin is :

X ____ 0 ____ 1 _____ 2 ______ 3 ______ 4

P(X):_0.0767_0.2762_0.3729_0.2238 _0.0503

Learn more :https://brainly.com/question/12474772

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