probability again, sum between 2 & 9?

Question

what is the probability of getting the sum between 2 and 9 when 2 die (6 sides) are rolled at the same time?

I assume that that "between 2 and 9" meaning greater than 2 and less than 9 so:

P( 2 and 9) = 26/36 - 1/36

prob. of getting the sum less than 9 - prob. of getting the sum of 2

but it doesn't seem right....? does anyone know how to solve this problem without listing and counting every outcome?

Actually I have an answer for that question already it's 25/36 but I don't know how to get that answer. That's why I'm asking you guys....

Answer 1

I'd assume between 2 and 9 would mean excluded 2 and 9.

So... you cannot get 2, 9, 10, 11 or 12.

Probability of 2 or 12 is 1/36 each
11 will be 2/36, 10 will be 3/36 and 9 will be 4/36

1 - 1/36 - 1/36 -2/36 - 3/36 - 4/36 = 25/36. Your original answer seems right.

Answer 2

a million) I worry that you made a mild mistake interior the first case. a short way of determining the fashion of strategies of having n with 2 dice is to undergo in ideas: n-a million for 2<=n<=6 13-n for 7<=n<=12 So for n = 6 there are 6-a million = 5 strategies - {(a million,5), (2,4), (3,3), (4,2), (5,a million)} and for n = 9 there are 13-9 = 4 strategies as you rightly reported - {(3, 6), (4, 5), (5, 4), 6,3)} If N is the progression that the the throw is n, then considering there are 6*6 = 36 strategies of throwing 2 dice (the size of the pattern area): P(N=6) = 5/36 and P(N=9) = 4/36 Now for 2 events A & B P(A or B) = P(A) + P(B) - P(A AND B) even with the indisputable fact that, if both events A & B are together unique, then P(A AND B) = 0 considering throwing a 6 or 9 are together unique (the end result won't be able to be both!), we've P(N=6 OR 9) = P(N=6) + P(N=9) - P(N = 6 AND 9) = 5/36 + 4/36 - 0 = 9/36 =a million/4 2) enable A be the progression that the the end result's a multiply of three. Then A = {3, 6, 9, 12} enable B be the progression that the throw is larger than 8. Then B = {9, 10, 11, 12} the following that's glaring that those 2 events are not any further together unique. For A AND B = {9, 12} If we denote by making use of N(X) the fashion of strategies for X to happen we come across that N(A) = 2+5+4+a million = 12 so as that P(A) = 12/36 N(B) = 4+3+2+a million = 10 so as that P(B) = 10/36 N(A AND B) = 4+a million = 5 so as that P(A AND B) = 5/36 subsequently P(A OR B) = P(A) + P(B) - P(A AND B) = 12/36 + 10/36 - 5/36 = 17/36 Voila!

Answer 3

I think you are wrong. Reading this problem, I would assume that two and nine are included.

There are 36 ways to roll the dice. There is one way to get 12, two ways to get 11 and three ways to get 10 -- so there are six ways to get 10 or higher. That leaves 30 ways to get 2-9. So -- the probability is 30/36 = 5/6.

In the unlikely event that you are right about the 2 & 9 not being included, there are four ways to get a nine and one way to get a two -- so the probability is 25/26.

If I were you, I'd list both answers.

Answer 4

it can be 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;4,1;
4,2,4,3;4,4;5,1;5,2;5,3;6,1;6,2
25 favourable outcomes and a sample space of 36
so probability=25/36

Answer 5

=)) not even wasting my time

do ur own homework

Answer 6

it can be12