You will need two 6-sided dice to complete this lab. The game: a pair of dice is

Question

You will need two 6-sided dice to complete this lab. The game: a pair of dice is rolled, and the outcome of the experiment is the sum of the dice.

- You win on the first roll if the sum is 7 or 11.

- You lose on the first roll if the sum is 2, 3, or 12.

- If the sum is 4, 5, 6, 8, 9, or 10, the number rolled is called the “point”.

Once the point is established, the rule is: If the player rolls a 7 before the point, the better loses; but if the point is rolled before a 7, the player wins

My outcome of the dice game:

Wins = 12 Loss = 8

Questions:

x = number of wins out of 20 trials (12)

n = 20

p = 0.4927

q = 0.5073

b. Using n = 20, and p = 0.4927:

- what is the probability that you would get exactly the number of wins that you achieved in the experiment in question 1?

- what is the probability that you would get your achieved number of wins or fewer?

- what is the probability that you would get your more than your achieved number of wins?

c. What is the expected number of wins in 20 trials?

Explanation / Answer

Solution:-

n = 20, p = 0.4927, q = 0.5073

1) The probability that you would get exactly the number of wins that you achieved in the experiment in question 1 is 0.1131

By applying binomial distribution:-

P(x, n, p) = nCx*px *(1 - p)(n - x)

P(x = 12) = 0.113077

2) The probability that you would get your achieved number of wins or fewer is 0.882.

By applying binomial distribution:-

P(x, n, p) = nCx*px *(1 - p)(n - x)

P(x < 12) = 0.88193

3) The probability that you would get your more than your achieved number of wins is 0.1181.

By applying binomial distribution:-

P(x, n, p) = nCx*px *(1 - p)(n - x)

P(x > 12) = 0.118065

c) The expected number of wins in 20 trials is 9.854

E(x) = n × p

E(x) = 20 × 0.4927

E(x) = 9.854

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