Question

6. (5) Rolling two four-sided dice, what’s the likelihood of getting the same result (i.e., the same sum) twice in a row?
7. (5) Rolling three four-sided dice, what is the probability of totaling at least 9?
8. (5) Rolling three four-sided dice twice, what is the expected total?
9. (5) Rolling three four-sided dice twice, what is the probability of actually getting exactly the expected total from Problem 8?
10. (10) Rolling three four-sided dice twice, what is the probability of the second total being larger than the first?

This is for a Decrete Structures class

Answer 1

a) if you roll a four sides dice each number from 1-4 will have equally likely outcome with probability 1/4

Now if you roll two dice total outcomes will be 4^2=16

Out of which same pairs are (1,1),(2,2),(3,3),(4,4)= total 4 times so the probability is 4/16 =1/4

b) if you are rolling three 4-sided dice then total outcomes are = 4*4*4=64

Now for sum of atleast 9 possible pairs are

(3,3,3),(3,3,4),(3,4,3),(4,3,3),(3,4,4),(4,3,4),(4,4,3),(4,4,4)

Total 8 times

So probability= 8/64= 1/8

c) expected total for rolling three 4-sided dice we have following possibilities

(1,1,1),(2,2,2),(3,3,3),(4,4,4)= 4

(1,1,2)*3!/2!= 3

(1,2,2)*3!/2!= 3

(1,2,3)*3!= 6

(1,1,3)*3!/2!= 3

(1,2,4)*3!=6

(1,3,4)*3!=6

(1,1,4)*3!/2!=3

(1,3,3)*3!/2!=3

(1,4,4)*3!/2!=3

(2,3,3)*3!/2!= 3

(2,3,4)*3!=6

(2,4,4)*3!/2!= 3

(3,3,4)*3!/2!=3

(3,4,4)*3!/2!=3

(1,3,4)*3!=6

Total 64 outcomes

Now there total are

3,6,9,12,8(6 times),11(3 times),10(3 times),10(3 times),9(6 times),8(3times),9(3 times),7(3times),6(3 times),8(6 times),7(6 times),5( 3 times),6 (6 times),5(3 times),4(3 times)

Therefore mean= 3+6+9+12+48+33+30+30+54+24+27+21+18+48+42+15+36+15+12/64

= 510/64= 7.96 ~apx 8

d) if we roll three 4-sided dice twice then total outcomes= 4*4*4*4*4*4=4^6= 4096

Now to get a sum of 8 possible outcomes

(1,1,1,1,2,2)*6!/4!*2!= 15 possible

(1,1,1,1,1,3)*6!/5!= 6 possible

Total 21 possibilities

Therefore probability= 21/4096

(I can answer first 4 parts only)