Question

Roll two 6-sided dice and examine their sum.

a) How long will it take to roll "snake-eyes" (a pair of ones)?
b) What is the probability of rolling a sum of 7 on the first roll?
c) What is the probability of rolling a sum of 7 on the fourth roll?
d) What is the probability of rolling a sum of 7 by the fourth roll?
e) What is the probability of it taking more than 10 rolls to roll the sum of 7?

Answer 1

a) For this one, I think you have to look at it from a geometric distribution point of view. (How many failures will it take to get a success, and your success is the pair of 1's). And the probability of this success (getting two 1's) is 1/36 But, we don't know how many trials are required to do so. So, I think the approach to take (I think I did a similar example once, I'll check my notes when I go home), is to consider:
You get it on the 1st try: 1/36
You get it on the 2nd try: 35/36 * 1/36 (you fail the 1st time, but you get it on the 2nd time)
3rd try: 35/36*35/36*1/36
....
until the N'th try. You must then add all these probabilities using an infinite geometric series I believe.

Answer 2

b) 6/36 = 1/6 (You can just enumerate all 6 poissbilities that gives you the sum "7")

c) First 3 rolls don't add to 7 with probability 1-6/36 = 30/36 for EACH roll. Thus you have
(30/36)^3 * 6/36

d) Some similar argument to a), you consider
1st try: 6/36
2nd try: 30/36 * 6/36
3rd try: (30/36)^2 * 6/36
4th: (30/36)^3 * 6/36
and you just add these probabilities together

Answer 3

do I add 1st, 2nd,3rd,4th to get the total probabilities

Answer 4

Yes for d)

Answer 5

ok

Answer 6

for a), I am doubting myself because when I calculated the probability, it just gave 1. But then I realized the question asks "How long it will take", rather than what is the probability... so um I am not too certain for a).

Answer 7

Although I guess you could said that you'd expect to roll about 36 times before get a snake-eyes

Answer 8

1.35

Answer 9

for e)
P(more than 10 rolls to get 7) = P(no 7's on the 1st 10 rolls).

We saw that the probability of of getting 7 = 1/6, so the probability of NOT getting 7 in a trial = 1 - 1/6 = 5/6

Hence, P(no 7s in 10 rolls) = (5/6)^10

Answer 10

for d) I got 0.5177

Answer 11

0.53

Answer 12

I added 0.17+0.14+0.12+0.10=0.53

Answer 13

oh I added the exact fractions, so this is just rounding error

Answer 14

ok, I re-added and got 0.53

Answer 15

so a is 1/36

Answer 16

No I think you should say it takes about 36 rolls to get a snake-eyes. Again, I was thinking of a probability at first, but they ask "How long will it take"

Answer 17

It looks like d and c is the same question

Answer 18

I see d say by

Answer 19

They look the same, but they aren't. The "on" and "by" make a difference.

in c), it asks the probability ON the 4th roll. This mean, you cannot get a "7" on roll 1, AND can';t get 7 on roll 2, AND can't get 7 on roll 4. Finally, you DO get a 7 on roll 4.

In d), it asks BY the 4th roll. This mean's, you can get a 7 on roll 1, OR a 7 on roll 2, OR a 7 on roll 3, OR a 7 on roll 4 (hence, you use the

Answer 20

hence you use the addition rule for each roll probability in d)

Answer 21

in c), it asks the probability ON the 4th roll. This mean, you cannot get a "7" on roll 1, AND can';t get 7 on roll 2, AND can't get 7 on roll 3. ** sorry for the typo.

Answer 22

That's ok and thank you

Answer 23

:)