1. Probability (7 points) problem, clearly describe the sample space and the ran
ID: 3887844 • Letter: 1
Question
1. Probability (7 points) problem, clearly describe the sample space and the random For the following variables you use. Be sure to justify where you get your expected values from. Consider playing a game where you roll n fair six-sided dice. For every 1 or 6 you roll you win $30, for rolling any other number you lose S3n (1) First assume n = 1 (i.e., you only roll one six-sided die) (a) (1 point) Describe the sample space for this experiment (b) (1 point) Describe a random variable which maps an outcome of this experiment to the winnings you receive (c) (1 point) Compute the expected value of this random variable (2) Now assume n = 6 (ie, you roll six six-sided dice) (a) (1 point) Describe the sample space for this experiment (you don't need to list the elements but describe what is contained in it) (b) (1 point) Describe a random variable which maps an outcome of this experiment to the winnings you receive. (Hint: Express your random variable as the sum of six random variables.) (c) (1 point) Compute the expected value of this random variable using the linearity of expectation. Based on this would you play this game? (3) (1 point) What is the largest value of n for which you would still want to play the game? Justify your answerExplanation / Answer
(1) n=1
Then the sample space for n=1 given by,
S={1,2,3,4,5,6}
hence n(S) = 6 i.e the total number of values
a) Sample space for this experiment given by,
S={1,2,3,4,5,6}
b)
let A be the sample space for winning outcome which is given by,
A={1,6}
c)
the probability distribution for X = amount won or lost is
X +30 -3
probability 2/6 4/6
expected value = 30*2/6 + -3*4/6 = 8
2) n=6
a)
if we roll a single die then we get 6 combination,
if we rolled 2 dice then we get 36 combination
and if we roll six six-side dice then we get 46,656 combination like (1,1,1,1,1,1),
(2,2,2,2,2,2,).
b)
Let B be the sample space for winning outcomes,
near about 500 we will get.
3)
Let we have only 1 roll.
expected payoff will be Each roll is equally likely, so it will show 1,2,3,4,5,61,2,3,4,5,6 with equal probability. Thus their average of 3.53.5 is the expected payoff.
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