For the following problem, clearly describe the sample space and the random vari

Question

For the following problem, clearly describe the sample space and the random variables you use. Be sure to justify where you get your expected values from. Consider playing a game where you roll k fair six-sided dice. For every 6 you roll you win $6, for rolling any other number you lose $1. First assume k = 1 (i.e., you only roll one six-sided die). Describe the sample space for this experiment. Describe a random variable which maps an outcome of this experiment to the winnings you receive. Compute the expected value of this random variable. Now assume k = 2 (i.e., you roll two six-sided dice). Describe the sample space for this experiment. Describe a random variable which maps an outcome of this experiment to the winnings you receive. Compute the expected value of t his random variable using the linearity of expectation. Based on this would you play this game?

Explanation / Answer

1 a)The sample space regarding the outcome of throwing one dice

is {1 ,2 ,3 ,4 ,5 ,6}

however if the sample space is needed according to the winnings then there are only two cases

case 1 . you get a six and win $ 6

case 2 . you do not get a six and lose $ 1

b)let X be the variable which maps the outcome of experiment to winnings

as we know probability of getting six is p(getting 6) = favourable outcome / sample apace

= 1/6= probability of winning $6

we know probability of not getting six is p(not getting 6) = favourable outcome / sample apace

= 5/6=probability of losing $1

so the random variable X can be mapped as

c)E(X) = $6 * 1/6 + (-$1)*5/6

= $1 -$ 5/6

= $1/6

2)a) throwing two fair dice can have 36 outcomes

{1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6}

but regarding winning there can be three outcomes or sample spaces

case 1) both having six as outcome and winning $6 +$6 = $12

case 2)one having 6 and other not having six as outcome and winning $6 - $1=$5

case 3) both dice not having 6 as outcome and losing $1 + $1= $2 (can also say winning -$2)

X $6 -$1 Probability 1/6 5/6

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