1. (8 points) Assume that you toss a fair six-faced die two times. (a) (1 pt) Ho

Question

1. (8 points) Assume that you toss a fair six-faced die two times.   

(a) (1 pt) How many possible outcomes are in the sample space? Explain your answer.

(b) (1 pt) What is the probability that you get a number greater than 3 at the first toss? (Show work and write the answer in simplest fraction form)

(c) (2 pts) What is the probability that the product of the two tosses is at most 5? (Show work and write the answer in simplest fraction form)

(d)   (2 pts) What is the probability that the product of the two tosses is at most 5, given that you get a number greater than 3 in the first toss? (Show work and write the answer in simplest fraction form)

(e) (2 pts) If event A is “Getting a number greater than 3 in the first toss” and event B is “The product of two tosses is at most 5”. Are event A and event B independent? Use statistical concept and mathematical expression to justify your answer.

Explanation / Answer

a) total number of outcomes =number of outcome on first die*number of outcome on second die=6*6=36

b) as there are 3 possiblities out of 6 to have a number greater then 3.

therefore probability =3/6 =1/2

c)probability that the product of the two tosses is at most 5=P((1,1),(1,2),(1,3),(!,4),(1.5),(2,1),(2,2),(3,1),(4,1),(5,1)

=10 outcomes out of 36 =10/36 =5/18

d)probability of number greater then 3 in first toss and product at most 5 =P((4,1),(5,1)) =2/36 =1/18

therefore  probability that the product of the two tosses is at most 5, given that you get a number greater than 3 in the first toss =(1/18)/(1/2)=2/18 =1/9

e)as P(A)*P(B)=(1/2)*(5/18)=5/36

which is not equal to P(A and B) =1/18

therefore not independent.

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