Question

1. Suppose you have a fair 6-sided die with the numbers 1 through 6 on the sides and a fair 5-sided die with the numbers 1 through 5 on the sides. What is the probability that a roll of the six-sided die will produce a value larger than the roll of the five-sided die?

2. What is the expected number of rolls until a fair five-sided die rolls a 3? Justify your answer briefly.

Answer #1

1)

P( roll of the six-sided die will produce a value larger than the roll of the five-sided die)

=P(1 on 5 sided die and 2 or more on 6 sided die)+P(2 on 5 sided die and 3 or more on 6 sided die)+P(3 on 5 sided die and 4 or more on 6 sided die)+P(4 on 5 sided die and 5 or more on 6 sided die)+P(5 on 5 sided die and 6 on 6 sided die)

=(1/5)*(5/6)+(1/5)*(4/6)+(1/5)*(3/6)+(1/5)*(2/6)+(1/5)*(1/6)=15/30 =1/2

2)

since probability of rolling a 3 on a five-sided die p=1/5 is fixed and independent from trail to trail.

therefore number of rolls until a fair five-sided die rolls a 3 follows geometric distribution, with parameter p=1/5

expected number of rolls until a fair five-sided die rolls a 3
=1/p=1/(1/5) =**5**

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